#set document(title: "12.1 Sequences", author: "OpenStax / XYZ Homework")
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== 12.1#h(0.6em)Sequences

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Evaluate #math.equation(block: false, alt: "2 n plus 3")[$2 n + 3$] for the integers 1, 2, 3, and 4. #linebreak() 
If you missed this problem, review .

#solutionbox[
5, 7, 9, 11
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate #math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power n")[$attach(( −1 ), t: n)$] for the integers 1, 2, 3, and 4. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "−1 , 1 , −1 , 1")[$−1 , 1 , −1 , 1$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
If #math.equation(block: false, alt: "f open parenthesis n close parenthesis equals n squared plus 2 ,")[$f ( n ) = n^(2) + 2 ,$] find #math.equation(block: false, alt: "f open parenthesis 1 close parenthesis plus f open parenthesis 2 close parenthesis plus f open parenthesis 3 close parenthesis .")[$f ( 1 ) + f ( 2 ) + f ( 3 ) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
20
]
]


=== Write the First Few Terms of a Sequence

Let’s look at the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x")[$f ( x ) = 2 x$] and evaluate it for just the counting numbers.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  [#strong[#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x")[$f ( x ) = 2 x$]]], [],
  [#strong[#math.equation(block: false, alt: "x")[$x$]]], [#strong[#math.equation(block: false, alt: "2 x")[$2 x$]]],
  [1], [2],
  [2], [4],
  [3], [6],
  [4], [8],
  [5], [10],
  […], […],
))

If we list the function values in order as 2, 4, 6, 8, and 10, … we have a sequence. A #strong[sequence] is a function whose domain is the counting numbers.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Sequences]

A #strong[sequence] is a function whose domain is the counting numbers.
]

A sequence can also be seen as an ordered list of numbers and each number in the list is a #emph[term]. A sequence may have an infinite number of terms or a finite number of terms. Our sequence has three dots (ellipsis) at the end which indicates the list never ends. If the domain is the set of all counting numbers, then the sequence is an #strong[infinite sequence]. Its domain is all counting numbers and there is an infinite number of counting numbers.

#math.equation(block: true, alt: "2 , 4 , 6 , 8 , 10 , … ,")[$2 , 4 , 6 , 8 , 10 , "…" ,$]If we limit the domain to a finite number of counting numbers, then the sequence is a #strong[finite sequence]. If we use only the first four counting numbers, 1, 2, 3, 4 our sequence would be the finite sequence,

#math.equation(block: true, alt: "2 , 4 , 6 , 8")[$2 , 4 , 6 , 8$]Often when working with sequences we do not want to write out all the terms. We want more compact way to show how each term is defined. When we worked with functions, we wrote #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x")[$f ( x ) = 2 x$] and we said the expression #math.equation(block: false, alt: "2 x")[$2 x$] was the rule that defined values in the range. While a sequence is a function, we do not use the usual function notation. Instead of writing the function as #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x ,")[$f ( x ) = 2 x ,$] we would write it as #math.equation(block: false, alt: "a sub n equals 2 n .")[$a_(n) = 2 n .$] The #math.equation(block: false, alt: "a sub n")[$a_(n)$] is the #emph[n]th term of the sequence, the term in the #emph[n]th position where #emph[n] is a value in the domain. The formula for writing the #emph[n]th term of the sequence is called the #strong[general term] or formula of the sequence.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[General Term of a Sequence]

The #strong[general term] of the sequence is found from the formula for writing the #emph[n]th term of the sequence. The #emph[n]th term of the sequence, #emph[a#sub[n]], is the term in the #emph[n]th position where #emph[n] is a value in the domain.
]

When we are given the general term of the sequence, we can find the terms by replacing #emph[n] with the counting numbers in order. For #math.equation(block: false, alt: "a sub n equals 2 n ,")[$a_(n) = 2 n ,$]

#figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#strong[#math.equation(block: false, alt: "n")[$n$]]], [1], [2], [3], [4], [5], [#math.equation(block: false, alt: "a sub n")[$a_(n)$]]),
  [#strong[#math.equation(block: false, alt: "a sub n")[$a_(n)$]]], [#math.equation(block: false, alt: "2 times 1; 2")[$2 · 1 \ 2$]], [#math.equation(block: false, alt: "2 times 2; 4")[$2 · 2 \ 4$]], [#math.equation(block: false, alt: "2 times 3; 6")[$2 · 3 \ 6$]], [#math.equation(block: false, alt: "2 times 4; 8")[$2 · 4 \ 8$]], [#math.equation(block: false, alt: "2 times 5; 10")[$2 · 5 \ 10$]], [2#emph[n]],
))

#math.equation(block: true, alt: "a sub 1 ,, a sub 2 ,, a sub 3 ,, a sub 4 ,, a sub 5 , … ,, a sub n , …; 2 ,, 4 ,, 6 ,, 8 ,, 10 , …")[$a_(1) , & a_(2) , & a_(3) , & a_(4) , & a_(5) "," #h(0.2em) "…" "," & a_(n) "," #h(0.2em) "…" \ 2 , & 4 , & 6 , & 8 , & 10 "," #h(0.2em) "…" &$]To find the values of a sequence, we substitute in the counting numbers in order into the general term of the sequence.

#examplebox("Example 1")[][
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals 4 n minus 3 .")[$a_(n) = 4 n − 3 .$]

#solutionbox[
We substitute the values 1, 2, 3, 4, and 5 into the formula, #math.equation(block: false, alt: "a sub n equals 4 n minus 3 ,")[$a_(n) = 4 n − 3 ,$] in order.

#figure(figph[This figure shows three rows and five columns. The first row reads nth term equals 4 times n minus 3 written five times. The second row reads a sub 1 equals 4 times g times 1 minus 3, a sub 2 equals 4 times g times 2 minus 3, a sub 3 equals 4 times g times 3 minus 3, a sub 4 equals 4 times g times 4 minus 3, a sub 5 equals 4 times g times 5 minus 3. The third row reads, a sub 1 equals 1, a sub 2 equals 5, a sub 3 equals 9, a sub 4 equals 13, a sub 5 equals 17.], alt: "This figure shows three rows and five columns. The first row reads nth term equals 4 times n minus 3 written five times. The second row reads a sub 1 equals 4 times g times 1 minus 3, a sub 2 equals 4 times g times 2 minus 3, a sub 3 equals 4 times g times 3 minus 3, a sub 4 equals 4 times g times 4 minus 3, a sub 5 equals 4 times g times 5 minus 3. The third row reads, a sub 1 equals 1, a sub 2 equals 5, a sub 3 equals 9, a sub 4 equals 13, a sub 5 equals 17.", caption: none)

The first five terms of the sequence are 1, 5, 9, 13, and 17.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals 3 n minus 4 .")[$a_(n) = 3 n − 4 .$]

#solutionbox[
#math.equation(block: true, alt: "−1 , 2 , 5 , 8 , 11")[$−1 , 2 , 5 , 8 , 11$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals 2 n minus 5 .")[$a_(n) = 2 n − 5 .$]

#solutionbox[
#math.equation(block: true, alt: "−3 , −1 , 1 , 3 , 5")[$−3 , −1 , 1 , 3 , 5$]
]
]

For some sequences, the variable is an exponent.

#examplebox("Example 2")[][
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals 2 to the power n plus 1 .")[$a_(n) = 2^(n) + 1 .$]

#solutionbox[
We substitute the values 1, 2, 3, 4, and 5 into the formula, #math.equation(block: false, alt: "a sub n equals 2 to the power n plus 1 ,")[$a_(n) = 2^(n) + 1 ,$] in order.

#figure(figph[This figure shows three rows and five columns. The first row reads “nth term equals 2 to the nth power plus 1” written five times. The second row reads, “a sub 1 equals 2 times 1 plus 1, a sub 2 equals 2 to the power of 2 plus 1, a sub 3 equals 2 to the power 3 plus 1, a sub 4 equals 2 to the power of 4 plus 1, a sub 5 equals 2 to the power 5 plus 1”. The last row reads “a sub 1 equals 3, a sub 2 equals 5, a sub 3 equals 9, a sub 4 equals 17, a sub 5 equals 33”.], alt: "This figure shows three rows and five columns. The first row reads “nth term equals 2 to the nth power plus 1” written five times. The second row reads, “a sub 1 equals 2 times 1 plus 1, a sub 2 equals 2 to the power of 2 plus 1, a sub 3 equals 2 to the power 3 plus 1, a sub 4 equals 2 to the power of 4 plus 1, a sub 5 equals 2 to the power 5 plus 1”. The last row reads “a sub 1 equals 3, a sub 2 equals 5, a sub 3 equals 9, a sub 4 equals 17, a sub 5 equals 33”.", caption: none)

The first five terms of the sequence are 3, 5, 9, 17, and 33.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals 3 to the power n plus 4 .")[$a_(n) = 3^(n) + 4 .$]

#solutionbox[
#math.equation(block: true, alt: "7 , 13 , 31 , 85 , 247")[$7 , 13 , 31 , 85 , 247$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals 2 to the power n minus 5 .")[$a_(n) = 2^(n) − 5 .$]

#solutionbox[
#math.equation(block: true, alt: "−3 , −1 , 3 , 11 , 27")[$−3 , −1 , 3 , 11 , 27$]
]
]

It is not uncommon to see the expressions #math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power n")[$attach(( −1 ), t: n)$] or #math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power n plus 1")[$attach(( −1 ), t: n + 1)$] in the general term for a sequence. If we evaluate each of these expressions for a few values, we see that this expression alternates the sign for the terms.

#figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#strong[#math.equation(block: false, alt: "n")[$n$]]], [#strong[1]], [#strong[2]], [#strong[3]], [#strong[4]], [#strong[5]]),
  [#strong[#math.equation(block: false, alt: "open parenthesis minus 1 close parenthesis to the power n")[$attach(( − 1 ), t: n)$]]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power 1; −1")[$attach(( −1 ), t: 1) \  \ −1$]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis squared; 1")[$attach(( −1 ), t: 2) \  \ 1$]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis cubed; −1")[$attach(( −1 ), t: 3) \  \ −1$]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power 4; 1")[$attach(( −1 ), t: 4) \  \ 1$]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power 5; −1")[$attach(( −1 ), t: 5) \  \ −1$]],
  [#strong[#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power n plus 1")[$attach(( −1 ), t: n + 1)$]]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power 1 plus 1; 1")[$attach(( −1 ), t: 1 + 1) \  \ 1$]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power 2 plus 1; −1")[$attach(( −1 ), t: 2 + 1) \  \ −1$]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power 3 plus 1; 1")[$attach(( −1 ), t: 3 + 1) \  \ 1$]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power 4 plus 1; −1")[$attach(( −1 ), t: 4 + 1) \  \ −1$]], [#math.equation(block: false, alt: "open parenthesis −1 close parenthesis to the power 5 plus 1; 1")[$attach(( −1 ), t: 5 + 1) \  \ 1$]],
))

#math.equation(block: true, alt: "a sub 1 ,, a sub 2 ,, a sub 3 ,, a sub 4 ,, a sub 5 , … ,, a sub n , …; −1 ,, 1 ,, −1 ,, 1 ,, −1 …; 1 ,, −1 ,, 1 ,, −1 ,, 1 …")[$a_(1) , & a_(2) , & a_(3) , & a_(4) , & a_(5) "," #h(0.2em) "…" "," & a_(n) "," #h(0.2em) "…" \ −1 , & 1 , & −1 , & 1 , & −1 "…" &  \ 1 , & −1 , & 1 , & −1 , & #h(0.6em) 1 "…" &$]The terms in the next example will alternate signs as a result of the powers of #math.equation(block: false, alt: "−1 .")[$−1 .$]

#examplebox("Example 3")[][
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n n cubed .")[$a_(n) = attach(( −1 ), t: n) n^(3) .$]

#solutionbox[
We substitute the values 1, 2, 3, 4, and 5 into the formula, #math.equation(block: false, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n n cubed ,")[$a_(n) = attach(( −1 ), t: n) n^(3) ,$] in order.

#figure(figph[This figure shows three rows and five columns. The first row reads “nth term equals negative 1 to the nth power times n cubed” written five times. The second row reads a sub 1 equals negative 1 to the power of 1 times g times 1 cubed, a sub 2 equals negative 1 squared time g times 2 cubed, a sub 3 equals negative 1 cubed times g times 23 cubed, a sub 4 equals negative 1 to the power of 4 times g times 4 cubed, a sub 5 equals negative 1 to the power of 5 times g times 5 cubed. The last row reads, “a sub 1 equals negative 1, a sub 2 equals 8, a sub 3 equals negative 27, a sub 4 equals 64, and a sub 5 equals negative 125.], alt: "This figure shows three rows and five columns. The first row reads “nth term equals negative 1 to the nth power times n cubed” written five times. The second row reads a sub 1 equals negative 1 to the power of 1 times g times 1 cubed, a sub 2 equals negative 1 squared time g times 2 cubed, a sub 3 equals negative 1 cubed times g times 23 cubed, a sub 4 equals negative 1 to the power of 4 times g times 4 cubed, a sub 5 equals negative 1 to the power of 5 times g times 5 cubed. The last row reads, “a sub 1 equals negative 1, a sub 2 equals 8, a sub 3 equals negative 27, a sub 4 equals 64, and a sub 5 equals negative 125.", caption: none)

The first five terms of the sequence are #math.equation(block: false, alt: "−1 , 8 , −27 , 64 ,")[$−1 , 8 , −27 , 64 ,$] and #math.equation(block: false, alt: "−125 .")[$−125 .$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n n squared .")[$a_(n) = attach(( −1 ), t: n) n^(2) .$]

#solutionbox[
#math.equation(block: true, alt: "−1 , 4 , −9 , 16 , −25")[$−1 , 4 , −9 , 16 , −25$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n plus 1 n cubed .")[$a_(n) = attach(( −1 ), t: n + 1) n^(3) .$]

#solutionbox[
#math.equation(block: true, alt: "1 , −8 , 27 , −64 , 125")[$1 , −8 , 27 , −64 , 125$]
]
]


=== Find a Formula for the General Term (#emph[n]th Term) of a Sequence

Sometimes we have a few terms of a sequence and it would be helpful to know the general term or #emph[n]th term. To find the general term, we look for patterns in the terms. Often the patterns involve multiples or powers. We also look for a pattern in the signs of the terms.

#examplebox("Example 4")[][
Find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "4 , 8 , 12 , 16 , 20 , and so on")[$4 , 8 , 12 , 16 , 20 "," #h(0.2em) …$]#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A numerical sequence showing multiples of 4: 4, 8, 12, 16, 20, ...], alt: "A numerical sequence showing multiples of 4: 4, 8, 12, 16, 20, ...", caption: none)]),
  [], [#figure(figph[Numerical sequence ranging from 1 to n, indicated by n.], alt: "Numerical sequence ranging from 1 to n, indicated by n.", caption: none)],
  [We look for a pattern in the terms.], [#figure(figph[A mathematical sequence titled 'Terms:' showing the numbers 4, 8, 12, 16, 20, indicating an arithmetic progression where each term increases by 4.], alt: "A mathematical sequence titled 'Terms:' showing the numbers 4, 8, 12, 16, 20, indicating an arithmetic progression where each term increases by 4.", caption: none)],
  [The numbers are all multiples of 4.], [#figure(figph[Number pattern showing the result of multiplying four by consecutive whole numbers: 1, 2, 3, 4, 5, and so on to n. Demonstrates an arithmetic sequence.], alt: "Number pattern showing the result of multiplying four by consecutive whole numbers: 1, 2, 3, 4, 5, and so on to n. Demonstrates an arithmetic sequence.", caption: none)],
  [], [The general term of the sequence is #math.equation(block: false, alt: "a sub n equals 4 n .")[$a_(n) = 4 n .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "3 , 6 , 9 , 12 , 15 , …")[$3 , 6 , 9 , 12 , 15 , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals 3 n")[$a_(n) = 3 n$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "5 , 10 , 15 , 20 , 25 , …")[$5 , 10 , 15 , 20 , 25 , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals 5 n")[$a_(n) = 5 n$]
]
]

#examplebox("Example 5")[][
Find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "2 , −4 , 8 , −16 , 32 , …")[$2 , −4 , 8 , −16 , 32 , "…"$]#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical sequence displaying numbers 2, -4, 8, -16, 32, ..., indicating a pattern of multiplying by -2 for each subsequent term.], alt: "A mathematical sequence displaying numbers 2, -4, 8, -16, 32, ..., indicating a pattern of multiplying by -2 for each subsequent term.", caption: none)]),
  [], [#figure(figph[A numerical sequence or index labeled 'n:' followed by the numbers 1, 2, 3, 4, 5, and then an ellipsis leading to '...n', indicating a sequence of integers from 1 to n.], alt: "A numerical sequence or index labeled 'n:' followed by the numbers 1, 2, 3, 4, 5, and then an ellipsis leading to '...n', indicating a sequence of integers from 1 to n.", caption: none)],
  [We look for a pattern in the terms.], [#figure(figph[A sequence of terms is shown: 2, -4, 8, -16, 32, ... This is a geometric progression where each term is multiplied by -2 to get the next term.], alt: "A sequence of terms is shown: 2, -4, 8, -16, 32, ... This is a geometric progression where each term is multiplied by -2 to get the next term.", caption: none)],
  [The numbers are powers of 2. The signs are #linebreak() alternating, with even #math.equation(block: false, alt: "n")[$n$] negative.], [#figure(figph[A mathematical pattern illustrating an alternating series, with each term defined by the formula (-1)^(n+1) \* 2^n, where n represents the term number starting from 1.], alt: "A mathematical pattern illustrating an alternating series, with each term defined by the formula (-1)^(n+1) * 2^n, where n represents the term number starting from 1.", caption: none)],
  [], [The general term of the sequence is #math.equation(block: false, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n plus 1 2 to the power n .")[$a_(n) = attach(( −1 ), t: n + 1) 2^(n) .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "−3 , 9 , −27 , 81 , −243 , …")[$−3 , 9 , −27 , 81 , −243 "," #h(0.2em) "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n 3 to the power n")[$a_(n) = attach(( −1 ), t: n) 3^(n)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a general term for the sequence whose first five terms are shown

#math.equation(block: true, alt: "1 , −4 , 9 , −16 , 25 , …")[$1 , −4 , 9 , −16 , 25 "," #h(0.2em) "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n plus 1 n squared")[$a_(n) = attach(( −1 ), t: n + 1) n^(2)$]
]
]

#examplebox("Example 6")[][
Find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "the fraction 1 over 3 , the fraction 1 over 9 , the fraction 1 over 27 , the fraction 1 over 81 , the fraction 1 over 243 , …")[$frac(1, 3) , frac(1, 9) , frac(1, 27) , frac(1, 81) , frac(1, 243) , "…"$]#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical sequence showing fractions 1/3, 1/9, 1/27, 1/81, 1/243, indicating a geometric progression where each term is 1 divided by increasing powers of three.], alt: "A mathematical sequence showing fractions 1/3, 1/9, 1/27, 1/81, 1/243, indicating a geometric progression where each term is 1 divided by increasing powers of three.", caption: none)]),
  [], [#figure(figph[A series of numbers for n: 1, 2, 3, 4, 5, as n continues.], alt: "A series of numbers for n: 1, 2, 3, 4, 5, as n continues.", caption: none)],
  [We look for a pattern in the terms.], [#figure(figph[Terms: 1/3, 1/9, 1/27, 1/81, 1/243... a geometric progression where each term is one third of the previous one.], alt: "Terms: 1/3, 1/9, 1/27, 1/81, 1/243... a geometric progression where each term is one third of the previous one.", caption: none)],
  [The numerators are all 1.], [#figure(figph[Mathematical pattern: a sequence of fractions where the numerator is 1 and the denominator is 3 raised to a positive integer power (1/3^n).], alt: "Mathematical pattern: a sequence of fractions where the numerator is 1 and the denominator is 3 raised to a positive integer power (1/3^n).", caption: none)],
  [The denominators are powers of 3.], [The general term of the sequence is #math.equation(block: false, alt: "a sub n equals the fraction 1 over 3 to the power n .")[$a_(n) = frac(1, 3^(n)) .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "the fraction 1 over 2 , the fraction 1 over 4 , the fraction 1 over 8 , the fraction 1 over 16 , the fraction 1 over 32 , …")[$frac(1, 2) , frac(1, 4) , frac(1, 8) , frac(1, 16) , frac(1, 32) , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals the fraction 1 over 2 to the power n")[$a_(n) = frac(1, 2^(n))$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "the fraction 1 over 1 , the fraction 1 over 4 , the fraction 1 over 9 , the fraction 1 over 16 , the fraction 1 over 25 , …")[$frac(1, 1) , frac(1, 4) , frac(1, 9) , frac(1, 16) , frac(1, 25) , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals the fraction 1 over n squared")[$a_(n) = frac(1, n^(2))$]
]
]


=== Use Factorial Notation

Sequences often have terms that are products of consecutive integers. We indicate these products with a special notation called #emph[factorial notation]. For example,#math.equation(block: false, alt: "5 !")[$5 !$], read 5 factorial, means #math.equation(block: false, alt: "5 times 4 times 3 times 2 times 1 .")[$5 · 4 · 3 · 2 · 1 .$] The exclamation point is not punctuation here; it indicates the #strong[factorial notation].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Factorial Notation]

If #emph[n] is a positive integer, then #math.equation(block: false, alt: "n !")[$n !$] is

#math.equation(block: true, alt: "n ! equals n open parenthesis n minus 1 close parenthesis open parenthesis n minus 2 close parenthesis and so on")[$n ! = n ( n − 1 ) ( n − 2 ) …$]We define #math.equation(block: false, alt: "0 !")[$0 !$] as 1, so #math.equation(block: false, alt: "0 ! equals 1 .")[$0 ! = 1 .$]
]

The values of #math.equation(block: false, alt: "n !")[$n !$] for the first 5 positive integers are shown.

#math.equation(block: true, alt: "1 !, 2 !, 3 !, 4 !, 5 !; 1, 2 times 1, 3 times 2 times 1, 4 times 3 times 2 times 1, 5 times 4 times 3 times 2 times 1; 1, 2, 6, 24, 120")[$1 ! &  &  & #h(2em) 2 ! &  &  & #h(2em) 3 ! &  &  & #h(2em) 4 ! &  &  & #h(2em) 5 ! \ 1 &  &  & #h(2em) 2 ⋅ 1 &  &  & #h(2em) 3 ⋅ 2 ⋅ 1 &  &  & #h(2em) 4 ⋅ 3 ⋅ 2 ⋅ 1 &  &  & #h(2em) 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 \ 1 &  &  & #h(2em) 2 &  &  & #h(2em) 6 &  &  & #h(2em) 24 &  &  & #h(2em) 120$]#examplebox("Example 7")[][
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals the fraction 1 over n !")[$a_(n) = frac(1, n !)$].

#solutionbox[
We substitute the values 1, 2, 3, 4, 5 into the formula, #math.equation(block: false, alt: "a sub n equals the fraction 1 over n ! ,")[$a_(n) = frac(1, n !) ,$] in order.

#figure(figph[This figure shows four rows and five columns. The first row reads, “nth term equals one divided by n factorial” written five times. The second row reads “a sub 1 equals one divided by 1 factorial, a sub 2 equals 1 divided by 2 factorial, a sub 3 equals 1 divided by 3 factorial, a sub 4 equals 1 divided by 4 factorial, a sub 5 equals 1 divided by 5 factorial”. The third row reads “a sub 1 equals 1 divided 1”, “a sub 2 equals 1 divided by 2 times g times 1”, “a sub 3 equals 1 divided by 3 times g times 2 g times 1”, “a sub 4 equals 1 divided 4 times g times 3 times g times 2 times g times 1”, “a sub 5 equals 1 divided by 5 g times 4 times g times 3 times g times 2 times g times 1”, “a sub 1 equals 1, a sub 2 equals one-half”, “a sub 3 equals one-sixth”, “a sub 4 equals 1 divided by 24”, “a sub 5 equals 1 divided by 120”.], alt: "This figure shows four rows and five columns. The first row reads, “nth term equals one divided by n factorial” written five times. The second row reads “a sub 1 equals one divided by 1 factorial, a sub 2 equals 1 divided by 2 factorial, a sub 3 equals 1 divided by 3 factorial, a sub 4 equals 1 divided by 4 factorial, a sub 5 equals 1 divided by 5 factorial”. The third row reads “a sub 1 equals 1 divided 1”, “a sub 2 equals 1 divided by 2 times g times 1”, “a sub 3 equals 1 divided by 3 times g times 2 g times 1”, “a sub 4 equals 1 divided 4 times g times 3 times g times 2 times g times 1”, “a sub 5 equals 1 divided by 5 g times 4 times g times 3 times g times 2 times g times 1”, “a sub 1 equals 1, a sub 2 equals one-half”, “a sub 3 equals one-sixth”, “a sub 4 equals 1 divided by 24”, “a sub 5 equals 1 divided by 120”.", caption: none)

The first five terms of the sequence are #math.equation(block: false, alt: "1 , the fraction 1 over 2 , the fraction 1 over 6 , the fraction 1 over 24 , the fraction 1 over 120 .")[$1 , frac(1, 2) , frac(1, 6) , frac(1, 24) , frac(1, 120) .$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals the fraction 2 over n ! .")[$a_(n) = frac(2, n !) .$]

#solutionbox[
#math.equation(block: true, alt: "2 , 1 , the fraction 1 over 3 , the fraction 1 over 12 , the fraction 1 over 60")[$2 , 1 , frac(1, 3) , frac(1, 12) , frac(1, 60)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals the fraction 3 over n ! .")[$a_(n) = frac(3, n !) .$]

#solutionbox[
#math.equation(block: true, alt: "3 , the fraction 3 over 2 , the fraction 1 over 2 , the fraction 1 over 8 , the fraction 1 over 40")[$3 , frac(3, 2) , frac(1, 2) , frac(1, 8) , frac(1, 40)$]
]
]

When there is a fraction with factorials in the numerator and denominator, we line up the factors vertically to make our calculations easier.

#examplebox("Example 8")[][
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals the fraction open parenthesis n plus 1 close parenthesis ! over open parenthesis n minus 1 close parenthesis ! .")[$a_(n) = frac(( n + 1 ) !, ( n − 1 ) !) .$]

#solutionbox[
We substitute the values 1, 2, 3, 4, 5 into the formula, #math.equation(block: false, alt: "a sub n equals the fraction open parenthesis n plus 1 close parenthesis ! over open parenthesis n minus 1 close parenthesis ! ,")[$a_(n) = frac(( n + 1 ) !, ( n − 1 ) !) ,$] in order.

#figure(figph[This figure shows five columns and five rows. The first row shows the sequence “nth term equals n plus 1 times factorial divided by n minus 1 times factorial” written five times. The second row is “a sub 1 equals 1 plus 1 times factorial divided by 1 minus 1 times factorial”, “a sub 2 equals 2 plus 1 times factorial divided by 2 minus 1 times factorial”, “a sub 3 equals 3 plus 1 times factorial divided by 3 minus 1 times factorial”, “a sub 4 equals 4 plus 1 times factorial divided by 4 minus 1 times factorial”, “a sub 5 equals 5 plus 1 times factorial divided by 5 minus 1 times factorial”. The third row reads “a sub 1 equals 2 times factorial divided by 0 times factorial”, “a sub 2 equals 3 times factorial divided by 1 times factorial”, “a sub 3 equals 4 times factorial divided by 2 times factorial”, “a sub 3 equals 4 times factorial divided by 2 times factorial”, “a sub 4 equals 5 times factorial divided by 3 times factorial”, “a sub 5 equals 6 times factorial divided by 4 times factorial”. The fourth row reads, “a sub 1 equals 2 times g time 1 divided by 1”, “a sub 2 equals 3 times g times 2 times g times 1 divided by 1”, “a sub 3 equals 4 times g times 3 times g times 2 times g times 1 divided by 2 times g times 1”, “a sub 4 equals 5 times g times 4 times g times 3 times g times 2 times g times 1 divided by 3 g times 2 times g times 1”, and “a sub 5 equals 6 times g times 5 times g times 4 times g times 3 times g times 2 times g times 1 divided by 4 times g times 3 times g times 2 times g times 1”. The fifth row reads “a sub 1 equals 2”, “a sub 2 equals 6”, “a sub 3 equals 12”, “a sub 4 equals 20”, “a sub 5 equals 30”.], alt: "This figure shows five columns and five rows. The first row shows the sequence “nth term equals n plus 1 times factorial divided by n minus 1 times factorial” written five times. The second row is “a sub 1 equals 1 plus 1 times factorial divided by 1 minus 1 times factorial”, “a sub 2 equals 2 plus 1 times factorial divided by 2 minus 1 times factorial”, “a sub 3 equals 3 plus 1 times factorial divided by 3 minus 1 times factorial”, “a sub 4 equals 4 plus 1 times factorial divided by 4 minus 1 times factorial”, “a sub 5 equals 5 plus 1 times factorial divided by 5 minus 1 times factorial”. The third row reads “a sub 1 equals 2 times factorial divided by 0 times factorial”, “a sub 2 equals 3 times factorial divided by 1 times factorial”, “a sub 3 equals 4 times factorial divided by 2 times factorial”, “a sub 3 equals 4 times factorial divided by 2 times factorial”, “a sub 4 equals 5 times factorial divided by 3 times factorial”, “a sub 5 equals 6 times factorial divided by 4 times factorial”. The fourth row reads, “a sub 1 equals 2 times g time 1 divided by 1”, “a sub 2 equals 3 times g times 2 times g times 1 divided by 1”, “a sub 3 equals 4 times g times 3 times g times 2 times g times 1 divided by 2 times g times 1”, “a sub 4 equals 5 times g times 4 times g times 3 times g times 2 times g times 1 divided by 3 g times 2 times g times 1”, and “a sub 5 equals 6 times g times 5 times g times 4 times g times 3 times g times 2 times g times 1 divided by 4 times g times 3 times g times 2 times g times 1”. The fifth row reads “a sub 1 equals 2”, “a sub 2 equals 6”, “a sub 3 equals 12”, “a sub 4 equals 20”, “a sub 5 equals 30”.", caption: none)

The first five terms of the sequence are 2, 6, 12, 20, and 30.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals the fraction open parenthesis n minus 1 close parenthesis ! over open parenthesis n plus 1 close parenthesis ! .")[$a_(n) = frac(( n − 1 ) !, ( n + 1 ) !) .$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 1 over 2 , the fraction 1 over 6 , the fraction 1 over 12 , the fraction 1 over 20 , the fraction 1 over 30")[$frac(1, 2) , frac(1, 6) , frac(1, 12) , frac(1, 20) , frac(1, 30)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the first five terms of the sequence whose general term is #math.equation(block: false, alt: "a sub n equals the fraction n ! over open parenthesis n plus 1 close parenthesis ! .")[$a_(n) = frac(n !, ( n + 1 ) !) .$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 1 over 2 , the fraction 1 over 3 , the fraction 1 over 4 , the fraction 1 over 5 , the fraction 1 over 6")[$frac(1, 2) , frac(1, 3) , frac(1, 4) , frac(1, 5) , frac(1, 6)$]
]
]


=== Find the Partial Sum

Sometimes in applications, rather than just list the terms, it is important for us to add the terms of a sequence. Rather than just connect the terms with plus signs, we can use #strong[summation notation].

For example, #math.equation(block: false, alt: "a sub 1 plus a sub 2 plus a sub 3 plus a sub 4 plus a sub 5")[$a_(1) + a_(2) + a_(3) + a_(4) + a_(5)$] can be written as #math.equation(block: false, alt: "∑ i equals 1 5 a sub i .")[$display(limits(∑)^(5)_(i = 1) a_(i)) .$] We read this as “the sum of #emph[a] sub #emph[i] from #emph[i] equals one to five.” The symbol #math.equation(block: false, alt: "∑")[$display(∑)$] means to add and the #emph[i] is the index of summation. The 1 tells us where to start (initial value) and the 5 tells us where to end (terminal value).

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Summation Notation]

The sum of the first #emph[n] terms of a sequence whose #emph[n]th term is #math.equation(block: false, alt: "a sub n")[$a_(n)$] is written in summation notation as:

#math.equation(block: true, alt: "∑ i equals 1 n a sub i equals a sub 1 plus a sub 2 plus a sub 3 plus a sub 4 plus a sub 5 plus and so on plus a sub n")[$display(limits(∑)^(n)_(i = 1) a_(i)) = a_(1) + a_(2) + a_(3) + a_(4) + a_(5) + … + a_(n)$]The #emph[i] is the index of summation and the 1 tells us where to start and the #emph[n] tells us where to end.
]

When we add a finite number of terms, we call the sum a #strong[partial sum].

#examplebox("Example 9")[][
Expand the partial sum and find its value: #math.equation(block: false, alt: "∑ i equals 1 5 2 i .")[$display(limits(∑)^(5)_(i = 1) 2 i) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "∑ i equals 1 5 2 i")[$display(limits(∑)^(5)_(i = 1) 2 i)$]]),
  [We substitute the values 1, 2, 3, 4, 5 in order.], [#math.equation(block: false, alt: "2 times 1 plus 2 times 2 plus 2 times 3 plus 2 times 4 plus 2 times 5")[$2 · 1 + 2 · 2 + 2 · 3 + 2 · 4 + 2 · 5$]],
  [Simplify.], [#math.equation(block: false, alt: "2 plus 4 plus 6 plus 8 plus 10")[$2 + 4 + 6 + 8 + 10$]],
  [Add.], [#math.equation(block: false, alt: "30")[$30$]],
  [], [#math.equation(block: false, alt: "∑ i equals 1 5 2 i equals 30")[$display(limits(∑)^(5)_(i = 1) 2 i) = 30$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Expand the partial sum and find its value: #math.equation(block: false, alt: "∑ i equals 1 5 3 i .")[$display(limits(∑)^(5)_(i = 1) 3 i) .$]

#solutionbox[
45
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Expand the partial sum and find its value: #math.equation(block: false, alt: "∑ i equals 1 5 4 i .")[$display(limits(∑)^(5)_(i = 1) 4 i) .$]

#solutionbox[
60
]
]

The index does not always have to be #emph[i] we can use any letter, but #emph[i] and #emph[k] are commonly used. The index does not have to start with 1 either—it can start and end with any positive integer.

#examplebox("Example 10")[][
Expand the partial sum and find its value: #math.equation(block: false, alt: "∑ k equals 0 3 the fraction 1 over k ! .")[$display(limits(∑)^(3)_(k = 0) frac(1, k !)) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "∑ k equals 0 3 the fraction 1 over k !")[$display(limits(∑)^(3)_(k = 0) frac(1, k !))$]]),
  [We substitute the values 0, 1, 2, 3, in order.], [#math.equation(block: false, alt: "the fraction 1 over 0 ! plus the fraction 1 over 1 ! plus the fraction 1 over 2 ! plus the fraction 1 over 3 !")[$frac(1, 0 !) + frac(1, 1 !) + frac(1, 2 !) + frac(1, 3 !)$]],
  [Evaluate the factorials.], [#math.equation(block: false, alt: "the fraction 1 over 1 plus the fraction 1 over 1 plus the fraction 1 over 2 plus the fraction 1 over 6")[$frac(1, 1) + frac(1, 1) + frac(1, 2) + frac(1, 6)$]],
  [Simplify.], [#math.equation(block: false, alt: "1 plus 1 plus the fraction 3 over 6 plus the fraction 1 over 6")[$1 + 1 + frac(3, 6) + frac(1, 6)$]],
  [Simplify.], [#math.equation(block: false, alt: "the fraction 16 over 6")[$frac(16, 6)$]],
  [Simplify.], [#math.equation(block: false, alt: "the fraction 8 over 3")[$frac(8, 3)$]],
  [], [#math.equation(block: false, alt: "∑ k equals 0 3 the fraction 1 over k ! equals the fraction 8 over 3")[$display(limits(∑)^(3)_(k = 0) frac(1, k !)) = frac(8, 3)$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Expand the partial sum and find its value: #math.equation(block: false, alt: "∑ k equals 0 3 the fraction 2 over k ! .")[$display(limits(∑)^(3)_(k = 0) frac(2, k !)) .$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 16 over 3")[$frac(16, 3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Expand the partial sum and find its value: #math.equation(block: false, alt: "∑ k equals 0 3 the fraction 3 over k ! .")[$display(limits(∑)^(3)_(k = 0) frac(3, k !)) .$]

#solutionbox[
8
]
]


=== Use Summation Notation to Write a Sum

In the last two examples, we went from summation notation to writing out the sum. Now we will start with a sum and change it to summation notation. This is very similar to finding the general term of a sequence. We will need to look at the terms and find a pattern. Often the patterns involve multiples or powers.

#examplebox("Example 11")[][
Write the sum using summation notation: #math.equation(block: false, alt: "1 plus the fraction 1 over 2 plus the fraction 1 over 3 plus the fraction 1 over 4 plus the fraction 1 over 5 .")[$1 + frac(1, 2) + frac(1, 3) + frac(1, 4) + frac(1, 5) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "1 plus the fraction 1 over 2 plus the fraction 1 over 3 plus the fraction 1 over 4 plus the fraction 1 over 5")[$#h(5.5em) 1 + frac(1, 2) + frac(1, 3) + frac(1, 4) + frac(1, 5)$]]),
  [], [#math.equation(block: false, alt: "n : 1 , 2 , 3 , 4 , 5")[$#h(1.5em) n ":" #h(1em) 1 , 2 , 3 , 4 , 5$]],
  [We look for a pattern in the terms.], [Terms: #math.equation(block: false, alt: "1 , the fraction 1 over 2 , the fraction 1 over 3 , the fraction 1 over 4 , the fraction 1 over 5")[$1 , frac(1, 2) , frac(1, 3) , frac(1, 4) , frac(1, 5)$]],
  [The numerators are all one.], [Pattern: #math.equation(block: false, alt: "the fraction 1 over 1 , the fraction 1 over 2 , the fraction 1 over 3 , the fraction 1 over 4 , the fraction 1 over 5 , and so on the fraction 1 over n")[$frac(1, 1) , frac(1, 2) , frac(1, 3) , frac(1, 4) , frac(1, 5) , … frac(1, n)$]],
  [The denominators are the counting numbers #linebreak() from one to five.], [The sum written in summation notation is],
  [], [#math.equation(block: false, alt: "1 plus the fraction 1 over 2 plus the fraction 1 over 3 plus the fraction 1 over 4 plus the fraction 1 over 5 equals ∑ n equals 1 5 the fraction 1 over n .")[$1 + frac(1, 2) + frac(1, 3) + frac(1, 4) + frac(1, 5) = display(limits(∑)^(5)_(n = 1) frac(1, n)) .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the sum using summation notation: #math.equation(block: false, alt: "the fraction 1 over 2 plus the fraction 1 over 4 plus the fraction 1 over 8 plus the fraction 1 over 16 plus the fraction 1 over 32 .")[$frac(1, 2) + frac(1, 4) + frac(1, 8) + frac(1, 16) + frac(1, 32) .$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 5 the fraction 1 over 2 to the power n")[$display(limits(∑)^(5)_(n = 1) frac(1, 2^(n)))$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write the sum using summation notation: #math.equation(block: false, alt: "1 plus the fraction 1 over 4 plus the fraction 1 over 9 plus the fraction 1 over 16 plus the fraction 1 over 25 .")[$1 + frac(1, 4) + frac(1, 9) + frac(1, 16) + frac(1, 25) .$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 5 the fraction 1 over n squared")[$display(limits(∑)^(5)_(n = 1) frac(1, n^(2)))$]
]
]

When the terms of a sum have negative coefficients, we must carefully analyze the pattern of the signs.

#examplebox("Example 12")[][
Write the sum using summation notation: #math.equation(block: false, alt: "−1 plus 8 minus 27 plus 64 minus 125 .")[$−1 + 8 − 27 + 64 − 125 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical expression showing the alternating sum of the first five cubes: -1 + 8 - 27 + 64 - 125.], alt: "A mathematical expression showing the alternating sum of the first five cubes: -1 + 8 - 27 + 64 - 125.", caption: none)]),
  [], [#figure(figph[The image displays the variable 'n:' followed by the numbers 1, 2, 3, 4, 5, arranged horizontally in a sequence.], alt: "The image displays the variable 'n:' followed by the numbers 1, 2, 3, 4, 5, arranged horizontally in a sequence.", caption: none)],
  [We look for a pattern in the terms.], [#figure(figph[A mathematical sequence labeled 'Terms:' shows the numbers -1, 8, -27, 64, -125, representing alternating positive and negative perfect cubes (1^3, 2^3, 3^3, 4^3, 5^3).], alt: "A mathematical sequence labeled 'Terms:' shows the numbers -1, 8, -27, 64, -125, representing alternating positive and negative perfect cubes (1^3, 2^3, 3^3, 4^3, 5^3).", caption: none)],
  [The signs of the terms alternate, #linebreak() and the odd terms are negative.], [#figure(figph[The image shows a mathematical sequence: (-1)^n multiplied by n^3 for n = 1, 2, 3, 4, 5. This creates an alternating series of cubic numbers.], alt: "The image shows a mathematical sequence: (-1)^n multiplied by n^3 for n = 1, 2, 3, 4, 5. This creates an alternating series of cubic numbers.", caption: none)],
  [The numbers are the cubes of the #linebreak() counting numbers from one to five.], [#figure(figph[A mathematical sequence displaying the pattern (-1)^n ×⋅ n^3 for n ranging from 1 to 5, with the cubed number shown in red.], alt: "A mathematical sequence displaying the pattern (-1)^n ×⋅ n^3 for n ranging from 1 to 5, with the cubed number shown in red.", caption: none)],
  [], [#figure(figph[The image displays a mathematical pattern: (-1)^n \* n^3.], alt: "The image displays a mathematical pattern: (-1)^n * n^3.", caption: none)],
  [], [The sum written in summation notation is],
  [], [#math.equation(block: false, alt: "minus 1 plus 8 minus 27 plus 64 minus 125 equals ∑ n equals 1 5 open parenthesis minus 1 close parenthesis to the power n times n cubed")[$− 1 + 8 − 27 + 64 − 125 = display(limits(∑)^(5)_(n = 1) attach(( − 1 ), t: n) ⋅ n^(3))$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write each sum using summation notation: #math.equation(block: false, alt: "1 minus 4 plus 9 minus 16 plus 25 .")[$1 − 4 + 9 − 16 + 25 .$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 5 open parenthesis −1 close parenthesis to the power n plus 1 n squared")[$display(limits(∑)^(5)_(n = 1) attach(( −1 ), t: n + 1) n^(2))$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write each sum using summation notation: #math.equation(block: false, alt: "−2 plus 4 minus 6 plus 8 minus 10 .")[$−2 + 4 − 6 + 8 − 10 .$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 5 open parenthesis −1 close parenthesis to the power n 2 n")[$display(limits(∑)^(5)_(n = 1) attach(( −1 ), t: n) 2 n)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access this online resource for additional instruction and practice with sequences.

- #link("https://openstax.org/l/37serseqfindpat")[Series and Sequences-Finding Patterns]
]


=== Key Concepts

- #strong[Factorial Notation] #linebreak() 
If #emph[n] is a positive integer, then #math.equation(block: false, alt: "n !")[$n !$] is #linebreak() 
#math.equation(block: true, alt: "n ! equals n open parenthesis n minus 1 close parenthesis open parenthesis n minus 2 close parenthesis and so on open parenthesis 3 close parenthesis open parenthesis 2 close parenthesis open parenthesis 1 close parenthesis .")[$n ! = n ( n − 1 ) ( n − 2 ) … ( 3 ) ( 2 ) ( 1 ) .$] #linebreak() 
We define #math.equation(block: false, alt: "0 !")[$0 !$] as 1, so #math.equation(block: false, alt: "0 ! equals 1")[$0 ! = 1$]
- #strong[Summation Notation] #linebreak() 
The sum of the first #emph[n] terms of a sequence whose #emph[n]th term #math.equation(block: false, alt: "a sub n")[$a_(n)$] is written in summation notation as: #linebreak() 
#math.equation(block: true, alt: "∑ i equals 1 n a sub i equals a sub 1 plus a sub 2 plus a sub 3 plus a sub 4 plus a sub 5 plus and so on plus a sub n")[$display(limits(∑)^(n)_(i = 1) a_(i)) = a_(1) + a_(2) + a_(3) + a_(4) + a_(5) + … + a_(n)$] #linebreak() 
The #emph[i] is the index of summation and the 1 tells us where to start and the #emph[n] tells us where to end.


==== Practice Makes Perfect

#strong[Write the First Few Terms of a Sequence]

In the following exercises, write the first five terms of the sequence whose general term is given.

#math.equation(block: true, alt: "a sub n equals 2 n minus 7")[$a_(n) = 2 n − 7$]

#solutionbox[
#math.equation(block: true, alt: "−5 , −3 , −1 , 1 , 3")[$−5 , −3 , −1 , 1 , 3$]
]

#math.equation(block: true, alt: "a sub n equals 5 n minus 1")[$a_(n) = 5 n − 1$]

#math.equation(block: true, alt: "a sub n equals 3 n plus 1")[$a_(n) = 3 n + 1$]

#solutionbox[
#math.equation(block: true, alt: "4 , 7 , 10 , 13 , 16")[$4 , 7 , 10 , 13 , 16$]
]

#math.equation(block: true, alt: "a sub n equals 4 n plus 2")[$a_(n) = 4 n + 2$]

#math.equation(block: true, alt: "a sub n equals 2 to the power n plus 3")[$a_(n) = 2^(n) + 3$]

#solutionbox[
#math.equation(block: true, alt: "5 , 7 , 11 , 19 , 35")[$5 , 7 , 11 , 19 , 35$]
]

#math.equation(block: true, alt: "a sub n equals 3 to the power n minus 1")[$a_(n) = 3^(n) − 1$]

#math.equation(block: true, alt: "a sub n equals 3 to the power n minus 2 n")[$a_(n) = 3^(n) − 2 n$]

#solutionbox[
#math.equation(block: true, alt: "1 , 5 , 21 , 73 , 233")[$1 , 5 , 21 , 73 , 233$]
]

#math.equation(block: true, alt: "a sub n equals 2 to the power n minus 3 n")[$a_(n) = 2^(n) − 3 n$]

#math.equation(block: true, alt: "a sub n equals the fraction 2 to the power n over n squared")[$a_(n) = frac(2^(n), n^(2))$]

#solutionbox[
#math.equation(block: true, alt: "2 , 1 , the fraction 8 over 9 , 1 , the fraction 32 over 25")[$2 , 1 , frac(8, 9) , 1 , frac(32, 25)$]
]

#math.equation(block: true, alt: "a sub n equals the fraction 3 to the power n over n cubed")[$a_(n) = frac(3^(n), n^(3))$]

#math.equation(block: true, alt: "a sub n equals the fraction 4 n minus 2 over 2 to the power n")[$a_(n) = frac(4 n − 2, 2^(n))$]

#solutionbox[
#math.equation(block: true, alt: "1 , the fraction 3 over 2 , the fraction 5 over 4 , the fraction 7 over 8 , the fraction 9 over 16")[$1 , frac(3, 2) , frac(5, 4) , frac(7, 8) , frac(9, 16)$]
]

#math.equation(block: true, alt: "a sub n equals the fraction 3 n plus 3 over 3 to the power n")[$a_(n) = frac(3 n + 3, 3^(n))$]

#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n times 2 n")[$a_(n) = attach(( −1 ), t: n) · 2 n$]

#solutionbox[
#math.equation(block: true, alt: "−2 , 4 , −6 , 8 , −10")[$−2 , 4 , −6 , 8 , −10$]
]

#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n times 3 n")[$a_(n) = attach(( −1 ), t: n) · 3 n$]

#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n plus 1 n squared")[$a_(n) = attach(( −1 ), t: n + 1) n^(2)$]

#solutionbox[
#math.equation(block: true, alt: "1 , −4 , 9 , −16 , 25")[$1 , −4 , 9 , −16 , 25$]
]

#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n plus 1 n to the power 4")[$a_(n) = attach(( −1 ), t: n + 1) n^(4)$]

#math.equation(block: true, alt: "a sub n equals the fraction open parenthesis −1 close parenthesis to the power n plus 1 over n squared")[$a_(n) = frac(attach(( −1 ), t: n + 1), n^(2))$]

#solutionbox[
#math.equation(block: true, alt: "1 , minus the fraction 1 over 4 , the fraction 1 over 9 , minus the fraction 1 over 16 , the fraction 1 over 25")[$1 , − frac(1, 4) , frac(1, 9) , − frac(1, 16) , frac(1, 25)$]
]

#math.equation(block: true, alt: "a sub n equals the fraction open parenthesis −1 close parenthesis to the power n plus 1 over 2 n")[$a_(n) = frac(attach(( −1 ), t: n + 1), 2 n)$]

#strong[Find a Formula for the General Term (#emph[n]th Term) of a Sequence]

In the following exercises, find a general term for the sequence whose first five terms are shown.

#math.equation(block: true, alt: "8 , 16 , 24 , 32 , 40 , …")[$8 , 16 , 24 , 32 , 40 , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals 8 n")[$a_(n) = 8 n$]
]

#math.equation(block: true, alt: "7 , 14 , 21 , 28 , 35 , …")[$7 , 14 , 21 , 28 , 35 , "…"$]

#math.equation(block: true, alt: "6 , 7 , 8 , 9 , 10 , …")[$6 , 7 , 8 , 9 , 10 , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals n plus 5")[$a_(n) = n + 5$]
]

#math.equation(block: true, alt: "−3 , −2 , −1 , 0 , 1 , …")[$−3 , −2 , −1 , 0 , 1 , "…"$]

#math.equation(block: true, alt: "e cubed , e to the power 4 , e to the power 5 , e to the power 6 , e to the power 7 , …")[$e^(3) , e^(4) , e^(5) , e^(6) , e^(7) , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals e to the power n plus 2")[$a_(n) = e^(n + 2)$]
]

#math.equation(block: true, alt: "the fraction 1 over e squared , the fraction 1 over e , 1 , e , e squared , …")[$frac(1, e^(2)) , frac(1, e) , 1 , e , e^(2) , "…"$]

#math.equation(block: true, alt: "−5 , 10 , −15 , 20 , −25 , …")[$−5 , 10 , −15 , 20 , −25 , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n 5 n")[$a_(n) = attach(( −1 ), t: n) 5 n$]
]

#math.equation(block: true, alt: "−6 , 11 , −16 , 21 , −26 , …")[$−6 , 11 , −16 , 21 , −26 , "…"$]

#math.equation(block: true, alt: "−1 , 8 , −27 , 64 , −125 , …")[$−1 , 8 , −27 , 64 , −125 , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n n cubed")[$a_(n) = attach(( −1 ), t: n) n^(3)$]
]

#math.equation(block: true, alt: "2 , −5 , 10 , −17 , 26 , …")[$2 , −5 , 10 , −17 , 26 , "…"$]

#math.equation(block: true, alt: "−2 , 4 , −6 , 8 , −10 , …")[$−2 , 4 , −6 , 8 , −10 , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n 2 n")[$a_(n) = attach(( −1 ), t: n) 2 n$]
]

#math.equation(block: true, alt: "1 , −3 , 5 , −7 , 9 , …")[$1 , −3 , 5 , −7 , 9 , "…"$]

#math.equation(block: true, alt: "the fraction 1 over 4 , the fraction 1 over 16 , the fraction 1 over 64 , the fraction 1 over 256 , the fraction 1 over 1,024 , …")[$frac(1, 4) , frac(1, 16) , frac(1, 64) , frac(1, 256) , frac(1, 1","024) , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals the fraction 1 over 4 to the power n")[$a_(n) = frac(1, 4^(n))$]
]

#math.equation(block: true, alt: "the fraction 1 over 1 , the fraction 1 over 8 , the fraction 1 over 27 , the fraction 1 over 64 , the fraction 1 over 125 , …")[$frac(1, 1) , frac(1, 8) , frac(1, 27) , frac(1, 64) , frac(1, 125) , "…"$]

#math.equation(block: true, alt: "minus the fraction 1 over 2 , minus the fraction 2 over 3 , minus the fraction 3 over 4 , minus the fraction 4 over 5 , minus the fraction 5 over 6 , …")[$− frac(1, 2) , − frac(2, 3) , − frac(3, 4) , − frac(4, 5) , − frac(5, 6) , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals minus the fraction n over n plus 1")[$a_(n) = − frac(n, n + 1)$]
]

#math.equation(block: true, alt: "−2 , minus the fraction 3 over 2 , minus the fraction 4 over 3 , minus the fraction 5 over 4 , minus the fraction 6 over 5 , …")[$−2 , − frac(3, 2) , − frac(4, 3) , − frac(5, 4) , − frac(6, 5) , "…"$]

#math.equation(block: true, alt: "minus the fraction 5 over 2 , minus the fraction 5 over 4 , minus the fraction 5 over 8 , minus the fraction 5 over 16 , minus the fraction 5 over 32 , …")[$− frac(5, 2) , − frac(5, 4) , − frac(5, 8) , − frac(5, 16) , − frac(5, 32) , "…"$]

#solutionbox[
#math.equation(block: true, alt: "a sub n equals minus the fraction 5 over 2 to the power n")[$a_(n) = − frac(5, 2^(n))$]
]

#math.equation(block: true, alt: "4 , the fraction 1 over 2 , the fraction 4 over 27 , the fraction 4 over 64 , the fraction 4 over 125 , …")[$4 , frac(1, 2) , frac(4, 27) , frac(4, 64) , frac(4, 125) , "…"$]

#strong[Use Factorial Notation]

In the following exercises, using factorial notation, write the first five terms of the sequence whose general term is given.

#math.equation(block: true, alt: "a sub n equals the fraction 4 over n !")[$a_(n) = frac(4, n !)$]

#solutionbox[
#math.equation(block: true, alt: "4 , 2 , the fraction 2 over 3 , the fraction 1 over 6 , the fraction 1 over 30")[$4 , 2 , frac(2, 3) , frac(1, 6) , frac(1, 30)$]
]

#math.equation(block: true, alt: "a sub n equals the fraction 5 over n !")[$a_(n) = frac(5, n !)$]

#math.equation(block: true, alt: "a sub n equals 3 n !")[$a_(n) = 3 n !$]

#solutionbox[
#math.equation(block: true, alt: "3 , 6 , 18 , 72 , 360")[$3 , 6 , 18 , 72 , 360$]
]

#math.equation(block: true, alt: "a sub n equals 2 n !")[$a_(n) = 2 n !$]

#math.equation(block: true, alt: "a sub n equals open parenthesis 2 n close parenthesis !")[$a_(n) = ( 2 n ) !$]

#solutionbox[
#math.equation(block: true, alt: "2 , 24 , 720 , 40320 , 3628800")[$2 , 24 , 720 , 40320 , 3628800$]
]

#math.equation(block: true, alt: "a sub n equals open parenthesis 3 n close parenthesis !")[$a_(n) = ( 3 n ) !$]

#math.equation(block: true, alt: "a sub n equals the fraction open parenthesis n minus 1 close parenthesis ! over open parenthesis n close parenthesis !")[$a_(n) = frac(( n − 1 ) !, ( n ) !)$]

#solutionbox[
#math.equation(block: true, alt: "1 , the fraction 1 over 2 , the fraction 1 over 3 , the fraction 1 over 4 , the fraction 1 over 5")[$1 , frac(1, 2) , frac(1, 3) , frac(1, 4) , frac(1, 5)$]
]

#math.equation(block: true, alt: "a sub n equals the fraction n ! over open parenthesis n plus 1 close parenthesis !")[$a_(n) = frac(n !, ( n + 1 ) !)$]

#math.equation(block: true, alt: "a sub n equals the fraction n ! over n squared")[$a_(n) = frac(n !, n^("2"))$]

#solutionbox[
#math.equation(block: true, alt: "1 , the fraction 1 over 2 , the fraction 2 over 3 , the fraction 3 over 2 , the fraction 24 over 5")[$1 , frac(1, 2) , frac(2, 3) , frac(3, 2) , frac(24, 5)$]
]

#math.equation(block: true, alt: "a sub n equals the fraction n squared over n !")[$a_(n) = frac(n^("2"), n !)$]

#math.equation(block: true, alt: "a sub n equals the fraction open parenthesis n plus 1 close parenthesis ! over n squared")[$a_(n) = frac(( n + 1 ) !, n^(2))$]

#solutionbox[
#math.equation(block: true, alt: "2 , the fraction 3 over 2 , the fraction 8 over 3 , the fraction 15 over 2 , the fraction 144 over 5")[$2 , frac(3, 2) , frac(8, 3) , frac(15, 2) , frac(144, 5)$]
]

#math.equation(block: true, alt: "a sub n equals the fraction open parenthesis n plus 1 close parenthesis ! over 2 n")[$a_(n) = frac(( n + 1 ) !, 2 n)$]

#strong[Find the Partial Sum]

In the following exercises, expand the partial sum and find its value.

#math.equation(block: true, alt: "∑ i equals 1 5 i squared")[$display(limits(∑)^(5)_(i = 1) i^(2))$]

#solutionbox[
#math.equation(block: true, alt: "1 plus 4 plus 9 plus 16 plus 25 equals 55")[$1 + 4 + 9 + 16 + 25 = 55$]
]

#math.equation(block: true, alt: "∑ i equals 1 5 i cubed")[$display(limits(∑)^(5)_(i = 1) i^(3))$]

#math.equation(block: true, alt: "∑ i equals 1 6 open parenthesis 2 i plus 3 close parenthesis")[$display(limits(∑)^(6)_(i = 1) ( 2 i) + 3 )$]

#solutionbox[
#math.equation(block: true, alt: "5 plus 7 plus 9 plus 11 plus 13 plus 15 equals 60")[$5 + 7 + 9 + 11 + 13 + 15 = 60$]
]

#math.equation(block: true, alt: "∑ i equals 1 6 open parenthesis 3 i minus 2 close parenthesis")[$display(limits(∑)^(6)_(i = 1) ( 3 i) − 2 )$]

#math.equation(block: true, alt: "∑ i equals 1 4 2 to the power i")[$display(limits(∑)^(4)_(i = 1) 2^(i))$]

#solutionbox[
#math.equation(block: true, alt: "2 plus 4 plus 8 plus 16 equals 30")[$2 + 4 + 8 + 16 = 30$]
]

#math.equation(block: true, alt: "∑ i equals 1 4 3 to the power i")[$display(limits(∑)^(4)_(i = 1) 3^(i))$]

#math.equation(block: true, alt: "∑ k equals 0 3 the fraction 4 over k !")[$display(limits(∑)^(3)_(k = 0) frac(4, k !))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 4 over 1 plus the fraction 4 over 1 plus the fraction 4 over 2 plus the fraction 4 over 6 equals the fraction 32 over 3 equals 10 the fraction 2 over 3")[$frac(4, 1) + frac(4, 1) + frac(4, 2) + frac(4, 6) = frac(32, 3) = 10 frac(2, 3)$]
]

#math.equation(block: true, alt: "∑ k equals 0 4 minus the fraction 1 over k !")[$display(limits(∑)^(4)_(k = 0) − frac(1, k !))$]

#math.equation(block: true, alt: "∑ k equals 1 5 k open parenthesis k plus 1 close parenthesis")[$display(limits(∑)^(5)_(k = 1) k ( k + 1 ))$]

#solutionbox[
#math.equation(block: true, alt: "2 plus 6 plus 12 plus 20 plus 30 equals 70")[$2 + 6 + 12 + 20 + 30 = 70$]
]

#math.equation(block: true, alt: "∑ k equals 1 5 k open parenthesis 2 k minus 3 close parenthesis")[$display(limits(∑)^(5)_(k = 1) k ( 2 k − 3 ))$]

#math.equation(block: true, alt: "∑ n equals 1 5 the fraction n over n plus 1")[$display(limits(∑)^(5)_(n = 1) frac(n, n + 1))$]

#solutionbox[
#math.equation(block: true, alt: "the fraction 1 over 2 plus the fraction 2 over 3 plus the fraction 3 over 4 plus the fraction 4 over 5 plus the fraction 5 over 6 equals the fraction 71 over 20")[$frac(1, 2) + frac(2, 3) + frac(3, 4) + frac(4, 5) + frac(5, 6) = frac(71, 20)$]
]

#math.equation(block: true, alt: "∑ n equals 1 4 the fraction n over n plus 2")[$display(limits(∑)^(4)_(n = 1) frac(n, n + 2))$]

#strong[Use Summation Notation to write a Sum]

In the following exercises, write each sum using summation notation.

#math.equation(block: true, alt: "the fraction 1 over 3 plus the fraction 1 over 9 plus the fraction 1 over 27 plus the fraction 1 over 81 plus the fraction 1 over 243")[$frac(1, 3) + frac(1, 9) + frac(1, 27) + frac(1, 81) + frac(1, 243)$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 5 the fraction 1 over 3 to the power n")[$display(limits(∑)^(5)_(n = 1) frac(1, 3^(n)))$]
]

#math.equation(block: true, alt: "the fraction 1 over 4 plus the fraction 1 over 16 plus the fraction 1 over 64 plus the fraction 1 over 256")[$frac(1, 4) + frac(1, 16) + frac(1, 64) + frac(1, 256)$]

#math.equation(block: true, alt: "1 plus the fraction 1 over 8 plus the fraction 1 over 27 plus the fraction 1 over 64 plus the fraction 1 over 125")[$1 + frac(1, 8) + frac(1, 27) + frac(1, 64) + frac(1, 125)$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 5 the fraction 1 over n cubed")[$display(limits(∑)^(5)_(n = 1) frac(1, n^(3)))$]
]

#math.equation(block: true, alt: "the fraction 1 over 5 plus the fraction 1 over 25 plus the fraction 1 over 125 plus the fraction 1 over 625")[$frac(1, 5) + frac(1, 25) + frac(1, 125) + frac(1, 625)$]

#math.equation(block: true, alt: "2 plus 1 plus the fraction 2 over 3 plus the fraction 1 over 2 plus the fraction 2 over 5")[$2 + 1 + frac(2, 3) + frac(1, 2) + frac(2, 5)$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 5 the fraction 2 over n")[$display(limits(∑)^(5)_(n = 1) frac(2, n))$]
]

#math.equation(block: true, alt: "3 plus the fraction 3 over 2 plus 1 plus the fraction 3 over 4 plus the fraction 3 over 5 plus the fraction 1 over 2")[$3 + frac(3, 2) + 1 + frac(3, 4) + frac(3, 5) + frac(1, 2)$]

#math.equation(block: true, alt: "3 minus 6 plus 9 minus 12 plus 15")[$3 − 6 + 9 − 12 + 15$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 5 open parenthesis −1 close parenthesis to the power n plus 1 3 n")[$display(limits(∑)^(5)_(n = 1) attach(( −1 ), t: n + 1) 3 n)$]
]

#math.equation(block: true, alt: "−5 plus 10 minus 15 plus 20 minus 25")[$−5 + 10 − 15 + 20 − 25$]

#math.equation(block: true, alt: "−2 plus 4 minus 6 plus 8 minus 10 plus and so on plus 20")[$−2 + 4 − 6 + 8 − 10 + … + 20$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 10 open parenthesis −1 close parenthesis to the power n 2 n")[$display(limits(∑)^(10)_(n = 1) attach(( −1 ), t: n) 2 n)$]
]

#math.equation(block: true, alt: "1 minus 3 plus 5 minus 7 plus 9 plus and so on plus 21")[$1 − 3 + 5 − 7 + 9 + … + 21$]

#math.equation(block: true, alt: "14 plus 16 plus 18 plus 20 plus 22 plus 24 plus 26")[$14 + 16 + 18 + 20 + 22 + 24 + 26$]

#solutionbox[
#math.equation(block: true, alt: "∑ n equals 1 7 open parenthesis 2 n plus 12 close parenthesis")[$display(limits(∑)^(7)_(n = 1) ( 2 n + 12 ))$]
]

#math.equation(block: true, alt: "9 plus 11 plus 13 plus 15 plus 17 plus 19 plus 21")[$9 + 11 + 13 + 15 + 17 + 19 + 21$]


==== Writing Exercises

In your own words, explain how to write the terms of a sequence when you know the formula. Show an example to illustrate your explanation.

#solutionbox[
Answers will vary.
]

Which terms of the sequence are negative when the #emph[n]#super[th] term of the sequence is #math.equation(block: false, alt: "a sub n equals open parenthesis −1 close parenthesis to the power n open parenthesis n plus 2 close parenthesis ?")[$a_(n) = attach(( −1 ), t: n) ( n + 2 ) ?$]

In your own words, explain what is meant by #math.equation(block: false, alt: "n !")[$n !$] Show some examples to illustrate your explanation.

#solutionbox[
Answers will vary.
]

Explain what each part of the notation #math.equation(block: false, alt: "∑ k equals 1 12 2 k")[$display(limits(∑)^(12)_(k = 1) 2 k)$] means.


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This figure shows a table with four columns and six rows. The first row is the header row and labels each column, “I can”, “Confidently”, “With some help”, and “No I don’t get it!”. The first row in the second column reads, “Write the first few terms of a sequence”, the third row, first column reads, “Find a Formula for the nth Term of a Sequence”, the fourth row first column reads “Use Factorial Notation, the fifth row, first column reads, Find the partial sum”, and the last row, first column reads, “Use Summation Notation to write a Sum”. The remaining three columns and rows are blank.], alt: "This figure shows a table with four columns and six rows. The first row is the header row and labels each column, “I can”, “Confidently”, “With some help”, and “No I don’t get it!”. The first row in the second column reads, “Write the first few terms of a sequence”, the third row, first column reads, “Find a Formula for the nth Term of a Sequence”, the fourth row first column reads “Use Factorial Notation, the fifth row, first column reads, Find the partial sum”, and the last row, first column reads, “Use Summation Notation to write a Sum”. The remaining three columns and rows are blank.", caption: none)

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.