3.2 Arithmetic and Geometric Sequences

We now turn to the question of finding closed formulas for particular types of sequences.

How do we know this? For the recursive definition, we need to specify $a sub 0$ . Then we need to express $a_{n}$ in terms of $a sub n minus 1$ . If we call the first term $a$ , then $a sub 0 equals a$ . For the recurrence relation, by the definition of an arithmetic sequence, the difference between successive terms is some constant, say $d$ . So $a_{n} - a_{n - 1} = d$ , or in other words,

$a_{0} = a a_{n} = a_{n - 1} + d$

To find a closed formula, first write out the sequence in general:

$\begin{aligned} a_{0} & = a \\ a_{1} & = a_{0} + d = a + d \\ a_{2} & = a_{1} + d = a + d + d = a + 2 d \\ a_{3} & = a_{2} + d = a + 2 d + d = a + 3 d \\ ⋮ \end{aligned}$

We see that to find the $n$ th term, we need to start with $a$ and then add $d$ a bunch of times. In fact, add it $n$ times. Thus $a_{n} = a + d n$ .

Example 1

Find recursive definitions and closed formulas for the arithmetic sequences below. Assume the first term listed is $a sub 0$ .

$2, 5, 8, 11, 14, and so on$ .
$50, 43, 36, 29, and so on$ .

First we should check that these sequences really are arithmetic by taking differences of successive terms. Doing so will reveal the common difference $d$ .

$5 - 2 = 3$ , $8 - 5 = 3$ , etc. To get from each term to the next, we add three, so $d equals 3$ . The recursive definition is therefore $a_{n} = a_{n - 1} + 3$ with $a sub 0 equals 2$ . The closed formula is $a_{n} = 2 + 3 n$ .
Here the common difference is $minus 7$ , since we add $minus 7$ to 50 to get 43, and so on. Thus we have a recursive definition of $a_{n} = a_{n - 1} - 7$ with $a sub 0 equals 50$ . The closed formula is $a_{n} = 50 - 7 n$ .

What about sequences like $2, 6, 18, 54, and so on$ ? This is not arithmetic because the difference between terms is not constant. However, the ratio between successive terms is constant. We call such sequences geometric.

The recursive definition for the geometric sequence with initial term $a$ and common ratio $r$ is $a_{n} = a_{n - 1} \cdot r; a_{0} = a$ . To get the next term we multiply the previous term by $r$ . We can find the closed formula like we did for the arithmetic progression. Write

$\begin{aligned} a_{0} & = a \\ a_{1} & = a_{0} \cdot r \\ a_{2} & = a_{1} \cdot r = a_{0} \cdot r \cdot r = a_{0} \cdot r^{2} \\ ⋮ \end{aligned}$

We must multiply the first term $a$ by $r$ a number of times, $n$ times to be precise. We get $a_{n} = a \cdot r^{n}$ .

Example 2

Find the recursive and closed formula for the geometric sequences below. Again, the first term listed is $a sub 0$ .

$3, 6, 12, 24, 48, and so on$
$27, 9, 3, 1, 1 / 3, and so on$

Start by checking that these sequences really are geometric by dividing each term by its previous term. If this ratio really is constant, we will have found $r$ .

$6 / 3 equals 2$ , $12 / 6 equals 2$ , $24 / 12 equals 2$ , etc. Yes, to get from any term to the next, we multiply by $r equals 2$ . So the recursive definition is $a_{n} = 2 a_{n - 1}$ with $a sub 0 equals 3$ . The closed formula is $a_{n} = 3 \cdot 2^{n}$ .
The common ratio is $r equals 1 / 3$ . So the sequence has recursive definition $a_{n} = \frac{1}{3} a_{n - 1}$ with $a sub 0 equals 27$ and closed formula $a_{n} = 27 \cdot {\frac{1}{3}}^{n}$ .

In the examples and formulas above, we assumed that the initial term was $a sub 0$ . If your sequence starts with $a sub 1$ , you can easily find the term that would have been $a sub 0$ and use that in the formula. For example, if we want a formula for the sequence $2, 5, 8, and so on$ and insist that $2 equals a sub 1$ , then we can find $a_{0} = - 1$ (since the sequence is arithmetic with common difference 3, we have $a_{0} + 3 = a_{1}$ ). Then the closed formula will be $a_{n} = - 1 + 3 n$ .

Sums of Arithmetic and Geometric Sequences

Investigate!

Your neighborhood grocery store has a candy machine full of Skittles.

Suppose that the candy machine currently holds exactly 650 Skittles, and every time someone inserts a quarter, exactly 7 Skittles come out of the machine.
1. How many Skittles will be left in the machine after 20 quarters have been inserted?
2. Will there ever be exactly zero Skittles left in the machine? Explain.
What if the candy machine gives 7 Skittles to the first customer who put in a quarter, 10 to the second, 13 to the third, 16 to the fourth, etc. How many Skittles has the machine given out after 20 quarters are put into the machine?
Now, what if the machine gives 4 Skittles to the first customer, 7 to the second, 12 to the third, 19 to the fourth, etc. How many Skittles has the machine given out after 20 quarters are put into the machine?

Look at the sequence $open parenthesis T sub n close parenthesis sub n greater than or equal to 1$ which starts $1, 3, 6, 10, 15, and so on$ . These are called the triangular numbers since they represent the number of dots in an equilateral triangle (think of how you arrange 10 bowling pins: a row of 4 plus a row of 3 plus a row of 2 and a row of 1).

A sequence of four dot patters, showing the first four triangular numbers: the patterns contain 1, 3, 6, and 10 dots, arranged in triangles with base 1, 2, 3, and 4, respectively.

Is this sequence arithmetic? No, since $3 - 1 = 2$ and $6 - 3 = 3 \neq 2$ , so there is no common difference. Is the sequence geometric? No. $3 / 1 equals 3$ but $6 / 3 equals 2$ , so there is no common ratio. What to do?

Notice that the differences between terms do form an arithmetic sequence: $2, 3, 4, 5, 6, and so on$ . This means that the $n$ th term of the sequence $(T_{n})$ is the sum of the first $n$ terms in the sequence $1, 2, 3, 4, 5, and so on$ . We say that $(T_{n})$ is the sequence of partial sums of the sequence $1, 2, 3, and so on$ (partial sums because we are not taking the sum of all infinitely many terms).

This should become clearer if we write the triangular numbers like this:

$\begin{aligned} 1 & = 1 \\ 3 & = 1 + 2 \\ 6 & = 1 + 2 + 3 \\ 10 & = 1 + 2 + 3 + 4 \\ ⋮ & ⋮ \\ T_{n} & = 1 + 2 + 3 + \dots + n \end{aligned}$

If we know how to add up the terms of an arithmetic sequence, we could find a closed formula for a sequence whose differences are the terms of that arithmetic sequence. Consider how we could find the sum of the first 100 positive integers (that is, $T sub 100$ ). Instead of adding them in order, we regroup and add $1 + 100 = 101$ . The next pair to combine is $2 + 99 = 101$ . Then $3 + 98 = 101$ . Keep going. This gives 50 pairs which each add up to $101$ , so $T_{100} = 101 \cdot 50 = 5050$ .¹

In general, using this same sort of regrouping, we find that $T_{n} = \frac{n (n + 1)}{2}$ . Incidentally, this is exactly the same as $(\binom{n + 1}{2})$ , which makes sense if you think of the triangular numbers as counting the number of handshakes that take place at a party with $n plus 1$ people: the first person shakes $n$ hands, the next shakes an additional $n minus 1$ hands and so on.

The point of all of this is that some sequences, while not arithmetic or geometric, can be interpreted as the sequence of partial sums of arithmetic and geometric sequences. Luckily there are methods we can use to compute these sums quickly.

Summing Arithmetic Sequences: Reverse and Add

Here is a technique that allows us to quickly find the sum of an arithmetic sequence.

Example 3

Find the sum: $2 + 5 + 8 + 11 + 14 + \dots + 470$ .

The idea is to mimic how we found the formula for triangular numbers. If we add the first and last terms, we get 472. The second term and second-to-last term also add up to 472. To keep track of everything, we might express this as follows. Call the sum $S$ . Then,

$S =$	$2$	$+$	$5$	$+$	$8$	$+ \dots +$	$467$	$+$	470
$+ S =$	$470$	$+$	$467$	$+$	$464$	$+ \dots +$	$5$	$+$	2
$2 S equals$	$472$	$+$	$472$	$+$	$472$	$+ \dots +$	$472$	$+$	$472$

To find $2 S$ then we add 472 to itself a number of times. What number? We need to decide how many terms (summands) are in the sum. Since the terms form an arithmetic sequence, the $n$ th term in the sum (counting $2$ as the 0th term) can be expressed as $2 plus 3 n$ . If $2 + 3 n = 470$ then $n equals 156$ . So $n$ ranges from 0 to 156, giving 157 terms in the sum. This is the number of 472's in the sum for $2 S$ . Thus

$2 S = 157 \cdot 472 = 74104$

It is now easy to find $S$ :

$S = 74104 / 2 = 37052$

This will work for the sum of any arithmetic sequence. Call the sum $S$ . Reverse and add. This produces a single number added to itself many times. Find the number of times. Multiply. Divide by 2. Done.

Example 4

Find a closed formula for $6 + 10 + 14 + \dots + (4 n - 2)$ .

Again, we have a sum of an arithmetic sequence. How many terms are in the sequence? Clearly each term in the sequence has the form $4 k minus 2$ (as evidenced by the last term). For which values of $k$ though? To get 6, $k equals 2$ . To get $4 n minus 2$ take $k = n$ . So to find the number of terms, we must count the number of integers in the range $2, 3, and so on, n$ . This is $n minus 1$ . (There are $n$ numbers from 1 to $n$ , so one less if we start with 2.)

Now reverse and add:

$S =$	$6$	$+$	$10$	$+ \dots +$	$4 n minus 6$	$+$	$4 n minus 2$
$+ S =$	$4 n minus 2$	$+$	$4 n minus 6$	$+ \dots +$	$10$	$+$	6
$2 S equals$	$4 n plus 4$	$+$	$4 n plus 4$	$+ \dots +$	$4 n plus 4$	$+$	$4 n plus 4$

Since there are $n minus 1$ terms, we get

$2 S = (n - 1) (4 n + 4) so S = \frac{(n - 1) (4 n + 4)}{2}$

Besides finding sums, we can use this technique to find closed formulas for sequences we recognize as sequences of partial sums.

Example 5

Use partial sums to find a closed formula for $open parenthesis a sub n close parenthesis sub n greater than or equal to 0$ which starts $2, 3, 7, 14, 24, 37, and so on and so on$

First, if you look at the differences between terms, you get a sequence of differences: $1, 4, 7, 10, 13, and so on$ , which is an arithmetic sequence. Written another way:

$\begin{aligned} a_{0} & = 2 \\ a_{1} & = 2 + 1 \\ a_{2} & = 2 + 1 + 4 \\ a_{3} & = 2 + 1 + 4 + 7 \end{aligned}$

and so on. We can write the general term of $(a_{n})$ in terms of the arithmetic sequence as follows:

$a_{n} = 2 + 1 + 4 + 7 + 10 + \dots + (1 + 3 (n - 1))$

(we use $1 + 3 (n - 1)$ instead of $1 plus 3 n$ to get the indices to line up correctly; for $a sub 3$ we add up to 7, which is $1 + 3 (3 - 1)$ ).

We can reverse and add, but the initial 2 does not fit our pattern. This just means we need to keep the 2 out of the reverse part:

$a_{n} =$	$2$	$+$	$1$	$+$	$4$	$+ \dots +$	$1 + 3 (n - 1)$
$+ a_{n} =$	$2$	$+$	$1 + 3 (n - 1)$	$+$	$1 + 3 (n - 2)$	$+ \dots +$	$1$
$2 a sub n equals$	$4$	$+$	$2 + 3 (n - 1)$	$+$	$2 + 3 (n - 1)$	$+ \dots +$	$2 + 3 (n - 1)$

Not counting the first term (the 4) there are $n$ summands of $2 + 3 (n - 1) = 3 n - 1$ so the right-hand side becomes $4 + (3 n - 1) n$ .

Finally, solving for $a_{n}$ we get

$a_{n} = \frac{4 + (3 n - 1) n}{2}$

. Just to be sure, we check $a_{0} = \frac{4}{2} = 2$ , $a_{1} = \frac{4 + 2}{2} = 3$ , etc. We have the correct closed formula.

Summing Geometric Sequences: Multiply, Shift and Subtract

To find the sum of a geometric sequence, we cannot just reverse and add. Do you see why? The reason we got the same term added to itself many times is because there was a constant difference. So as we added that difference in one direction, we subtracted the difference going the other way, leaving a constant total. For geometric sums, we have a different technique.

To better see what happened in the above example, try writing it this way:

$S =$	$3 plus$	$6 + 12 + 24 + \dots + 12288$
$- 2 S =$		$6 + 12 + 24 + \dots + 12288$	$plus 24576$
$- S =$	$3 plus$	$0 + 0 + 0 + \dots + 0$	$minus 24576$

Then divide both sides by $minus 1$ and we have the same result for $S$ . The idea is, by multiplying the sum by the common ratio, each term becomes the next term. We shift over the sum to get the subtraction to mostly cancel out, leaving just the first term and new last term.

Example 7

Find a closed formula for $S (n) = 2 + 10 + 50 + \dots + 2 \cdot 5^{n}$ .

The common ratio is 5. So we have

$S$	$= 2 + 10 + 50 + \dots + 2 \cdot 5^{n}$
$minus 5 S$	$= 10 + 50 + \dots + 2 \cdot 5^{n} + 2 \cdot 5^{n + 1}$
$minus 4 S$	$= 2 - 2 \cdot 5^{n + 1}$

Thus $S = \frac{2 - 2 \cdot 5^{n + 1}}{- 4}$

Even though this might seem like a new technique, you have probably used it before.

Example 8

Express $0.464646 and so on$ as a fraction.

Let $N equals 0.46464646 and so on$ . Consider $0.01 N$ . We get:

$N =$	$0.4646464 and so on$
$- 0.01 N =$	$0.00464646 and so on$
$0.99 N equals$	$0.46$

So $N = \frac{46}{99}$ . What have we done? We viewed the repeating decimal $0.464646 and so on$ as a sum of the geometric sequence $0.46, 0.0046, 0.000046, and so on$ The common ratio is $0.01$ . The only real difference is that we are now computing an infinite geometric sum, we do not have the extra “last” term to consider. Really, this is the result of taking a limit as you would in calculus when you compute infinite geometric sums.

Is there a pair of integers $(a, b)$ such that $a, x sub 1, y sub 1, b$ is part of an arithmetic sequences and $a, x sub 2, y sub 2, b$ is part of a geometric sequence with $x sub 1, x sub 2, y sub 1, y sub 2$ all integers?

Consider the sequence $2, 7, 15, 26, 40, 57, and so on$ (with $a sub 0 equals 2$ ). By looking at the differences between terms, express the sequence as a sequence of partial sums. Then find a closed formula for the sequence by computing the $n$ th partial sum.

We have $2 equals 2$ , $7 = 2 + 5$ , $15 = 2 + 5 + 8$ , $26 = 2 + 5 + 8 + 11$ , and so on. The terms in the sums are given by the arithmetic sequence $b_{n} = 2 + 3 n$ . In other words, $a_{n} = \sum_{k = 0}^{n} (2 + 3 k)$ . To find the closed formula, we reverse and add. We get $a_{n} = \frac{(4 + 3 n) (n + 1)}{2}$ (we have $n plus 1$ there because there are $n plus 1$ terms in the sum for $a_{n}$ ).

Starting with any rectangle, we can create a new, larger rectangle by attaching a square to the longer side. For example, if we start with a $2 times 5$ rectangle, we would glue on a $5 times 5$ square, forming a $5 times 7$ rectangle:

The next rectangle would be formed by attaching a $7 times 7$ square to the top or bottom of the $5 times 7$ rectangle.

Create a sequence of rectangles using this rule starting with a $1 times 2$ rectangle. Then write out the sequence of perimeters for the rectangles (the first term of the sequence would be 6, since the perimeter of a $1 times 2$ rectangle is 6 - the next term would be 10).
Repeat the above part this time starting with a $1 times 3$ rectangle.
Find recursive formulas for each of the sequences of perimeters you found in parts (a) and (b). Don't forget to give the initial conditions as well.
Are the sequences arithmetic? Geometric? If not, are they close to being either of these (i.e., are the differences or ratios almost constant)? Explain.

If you have enough toothpicks, you can make a large triangular grid. Below, are the triangular grids of size 1 and of size 2. The size 1 grid requires 3 toothpicks, the size 2 grid requires 9 toothpicks.

Three toothpicks arranged as the sides of an equilateral triangle.

Nine toothpicks arranged into a triangle with two toothpicks forming each edge, and an upside-down triangle in the center.

Let $t_{n}$ be the number of toothpicks required to make a size $n$ triangular grid. Write out the first 5 terms of the sequence $t sub 1, t sub 2, and so on$ .
Find a recursive definition for the sequence. Explain why you are correct.
Is the sequence arithmetic or geometric? If not, is it the sequence of partial sums of an arithmetic or geometric sequence? Explain why your answer is correct.
Use your results from part (c) to find a closed formula for the sequence. Show your work.

If you were to shade in a $n \times n$ square on graph paper, you could do it the boring way (with sides parallel to the edge of the paper) or the interesting way, as illustrated below:

Five squares arranged as a plus sign. Viewed another way, the squares are arranged in three centered rows of 1, 3, and 1 squares.

13 squares arranged in five centered rows, containing 1, 3, 5, 3, and 1 square each.

25 squares arranged in rows of length 1, 3, 5, 7, 5, 3, and 1.

The interesting thing here, is that a $3 times 3$ square now has area 13. Our goal is the find a formula for the area of a $n \times n$ (diagonal) square.

Write out the first few terms of the sequence of areas (assume $a sub 1 equals 1$ , $a sub 2 equals 5$ , etc). Is the sequence arithmetic or geometric? If not, is it the sequence of partial sums of an arithmetic or geometric sequence? Explain why your answer is correct, referring to the diagonal squares.
Use your results from part (a) to find a closed formula for the sequence. Show your work. Note, while there are lots of ways to find a closed formula here, you should use partial sums specifically.
Find the closed formula in as many other interesting ways as you can.

Here is a surprising use of sequences to answer a counting question: How many license plates consist of 6 symbols, using only the three numerals 1, 2, and 3 and the four letters a, b, c, and d, so that no numeral appears after any letter? For example, “31ddac”, “123321”, and “ababab” are each acceptable license plates, but “13ba2c” is not.

First answer this question by considering different cases: how many of the license plates contain no numerals? How many contain one numeral, etc.
Now use the techniques of this section to show why the answer is $4^{7} - 3^{7}$ .

Discrete Mathematics: An Open Introduction, 3rd edition, by Oscar Levin (discrete.openmathbooks.org), licensed under CC BY-SA 4.0; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.