Rate of Growth

Section 4.2 Rate of Growth

Subsection Arithmetic and Geometric Sequences

Investigate!

For the patterns of dots below, draw the next pattern in the sequence. Describe the rate of growth of the number of dots in the pattern. Then give guess a recursive definition and a closed formula for the number of dots in the $n$th pattern.

$A sequence of three dot patterns, starting with a single dot (labeled n = 0), then the same dot with four new dots extending into an X (labeled n = 1) and finally the same pattern but with four new dots extending onto the limbs of the X (labeled n = 2).$

$A sequence of three dot patterns, labeled n = 0, n = 1, and n = 2, from left to right. The first pattern contains two dots, on above the other. The second pattern can be viewed as each of the dots from the first pattern splitting into 3, forming two triangles (so six dots total). The third patter again takes each dot from the second pattern and splits them into three, arranged as triangles (so 18 dots all together).$

$A sequence of four dot patters, labeled n = 1 through n = 4. The first pattern is a single dot. The second adds a row of two dots below the single dot (so three dots in a triangle). The third pattern adds a row of three dots, creating six dots arranged in a triangle. Finally we add four dots below the previous six, still forming a triangle, this time of 10 dots.$

Our goal is to find closed formulas for sequences. Our primary strategy will be to first determine how the sequence is changing from term to term. This will lead to a recurrence relation for the sequence, and from that recurrence relation, we will find a closed formula. We start with two types of sequences that are particularly common and useful: arithmetic and geometric sequences.

Suppose you start a business selling prints of mathematical art. In week zero, you sell one print. Each week after that, you sell four more prints than you did the previous week. How many prints will you sell in the $n$th week?

We can easily compute the first few terms of the sequence: $1, 5, 9, 13,\ldots\text{.}$ How do I know this is correct? From the problem, we see that to get from one term to the next, we must add 4. I is clear then that the recurrence relation for the sequence is

\begin{equation*} a_n = a_{n-1} + 4\text{.} \end{equation*}

The rate of growth for the sequence is the constant $4\text{,}$ since the difference between any two terms is 4 (note, we could write the recurrence relation as $a_n - a_{n-1} = 4$).

We call sequences with a constant rate of change arithmetic sequences.

Now let’s find a closed formula for our sequence. The first term is $a_0 = 1\text{.}$ To get $a_1\text{,}$ we add $4\text{.}$ The next term requires us to add $4$ again, which means we have added $4$ to our initial term twice. Then we add $4$ again, for a total of three times for $a_3\text{.}$ In fact, to get $a_n\text{,}$ we will have added $4$ to $a_0$ a total of $n$ times. Thus, the closed formula for the sequence is

\begin{equation*} a_n = 1 + 4n\text{.} \end{equation*}

This works for any arithmetic sequence. That is, any sequence with a constant difference will have a linear closed formula, where the “slope” of the linear function is the common difference.

Arithmetic Sequences.

If the terms of a sequence differ by a constant, we say the sequence is arithmetic. If the initial term ($a_0$) of the sequence is $a$ and the common difference is $d\text{,}$ then we have,

Recursive definition: $a_n = a_{n-1} + d$ with $a_0 = a\text{.}$

Closed formula: $a_n = a + dn\text{.}$

As we did for our example above, for the recursive definition, we need to specify $a_0\text{.}$ Then we need to express $a_n$ in terms of $a_{n-1}\text{.}$ If we call the first term $a\text{,}$ then $a_0 = a\text{.}$ For the recurrence relation, by the definition of an arithmetic sequence, the difference between successive terms is some constant, say $d\text{.}$ So $a_n - a_{n-1} = d\text{,}$ or in other words,

\begin{equation*} a_0 = a \qquad a_n=a_{n-1}+d\text{.} \end{equation*}

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

\begin{align*} a_0 \amp = a\\ a_1 \amp = a_0 + d = a+d\\ a_2 \amp = a_1 + d = a+d+d = a+2d\\ a_3 \amp = a_2 + d = a+2d+d = a+3d\\ \amp \vdots \end{align*}

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 $a_n = a+dn\text{.}$

Example 4.2.1.

Find recursive definitions and closed formulas for the arithmetic sequences below. Assume the first term listed is $a_0\text{.}$

$2, 5, 8, 11, 14, \ldots\text{.}$
$50, 43, 36, 29, \ldots\text{.}$

Solution.

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

$5-2 = 3\text{,}$ $8-5 = 3\text{,}$ etc. To get from each term to the next, we add three, so $d = 3\text{.}$ The recursive definition is therefore $a_n = a_{n-1} + 3$ with $a_0 = 2\text{.}$ The closed formula is $a_n = 2 + 3n\text{.}$
Here the common difference is $-7\text{,}$ since we add $-7$ to 50 to get 43, and so on. Thus we have a recursive definition of $a_n = a_{n-1} - 7$ with $a_0 = 50\text{.}$ The closed formula is $a_n = 50 - 7n\text{.}$

What about sequences like $2, 6, 18, 54, \ldots\text{?}$ 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 $a_n = a_{n-1}\cdot r; a_0 = a\text{.}$ To get the next term we multiply the previous term by $r\text{.}$ We can find the closed formula like we did for the arithmetic progression. Write

\begin{align*} a_0 \amp = a\\ a_1 \amp = a_0\cdot r\\ a_2 \amp = a_1 \cdot r = a_0\cdot r\cdot r = a_0\cdot r^2\\ \amp \vdots \end{align*}

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

Geometric Sequences.

A sequence is called geometric if the ratio between successive terms is constant. Suppose the initial term $a_0$ is $a$ and the common ratio is $r\text{.}$ Then we have,

Recursive definition: $a_n = ra_{n-1}$ with $a_0 = a\text{.}$

Closed formula: $a_n = a\cdot r^{n}\text{.}$

Example 4.2.2.

Find the recursive and closed formula for the geometric sequences below. Again, the first term listed is $a_0\text{.}$

$\displaystyle 3, 6, 12, 24, 48, \ldots$
$\displaystyle 27, 9, 3, 1, 1/3, \ldots$

Solution.

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 $r\text{.}$

$6/3 = 2\text{,}$ $12/6 = 2\text{,}$ $24/12 = 2\text{,}$ etc. Yes, to get from any term to the next, we multiply by $r = 2\text{.}$ So the recursive definition is $a_n = 2a_{n-1}$ with $a_0 = 3\text{.}$ The closed formula is $a_n = 3\cdot 2^{n}\text{.}$
The common ratio is $r = 1/3\text{.}$ So the sequence has recursive definition $a_n = \frac{1}{3}a_{n-1}$ with $a_0 = 27$ and closed formula $a_n = 27\cdot \frac{1}{3}^{n}\text{.}$

Geometric sequences are those which have a growth rate that is proportional to the sequence itself. Just like you might have seen in Calculus, it is exactly the exponential functions that have this property.

In the examples and formulas above, we assumed that the initial term was $a_0\text{.}$ If your sequence starts with $a_1\text{,}$ you can easily find the term that would have been $a_0$ and use that in the formula. For example, if we want a formula for the sequence $2, 5, 8,\ldots$ and insist that $2= a_1\text{,}$ 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 $a_n = -1 + 3n\text{.}$

Remark 4.2.3.

If you look at other textbooks or online, you might find that their closed formulas for arithmetic and geometric sequences differ from ours. Specifically, you might find the formulas $a_n = a +(n-1)d$ (arithmetic) and $a_n = a\cdot r^{n-1}$ (geometric). Which is correct? Both! In our case, we take $a$ to be $a_0\text{.}$ If instead we had $a_1$ as our initial term, we would get the (slightly more complicated) formulas you find elsewhere.

Reading Questions Reading Questions

1.

Match each formula on the left with the type of formula described on the right.

$a_n = 2n+3$
Arithmetic; closed
$a_n = a_{n-1} + 3$
Arithmetic; recursive
$a_n = 3\cdot 2^n$
Geometric; closed
$a_n = 3a_{n-1}$
Geometric; recursive

2.

How can you decide whether a sequence is the sequence of partial sums of a arithmetic or geometric sequence? Describe what you would do to check, using an example.

Exercises Practice Problems

1.

Consider the sequence $6, 12, 18, 24, 30, \ldots$ with $a_1 = 6\text{.}$

Which of the following is a recursive definition for the sequence. Select all that apply.
- $a_n = a_{n-1} + a_{n-2}$; $a_1 = 6$
- $a_n = 6 \cdot a_{n-1}$; $a_1 = 6$
- $a_n = a_{n-1} + 6$; $a_1 = 6$
- $a_n = 6 \cdot 6^n$
Give a closed formula for the $n$th term of the sequence.

$a_n =$
Is 3102 a term in the sequence?
- No, it is between 3100 and 3106
- Yes, it is $a_{516}$
- Yes, it is $a_{517}$
- No, it is larger than 822
- Yes, it is $a_{3102}$
How many terms does the finite sequence $6, 12, 18, 24, 30, \ldots, {822}$ have?
Find the sum: $6 + 12 + 18 + 24 + 30 + \cdots + {822}$
Use what you found above to find $b_n\text{,}$ the $n$th term of $1, {7}, {19}, {37}, {61}, \ldots$ where $b_0 = 1\text{.}$

$b_n =$

Solution.

This is an arithmetic sequence with initial term 6 and common difference 6.

For the last part, you are finding the sequence of partial sums, added to 1.

2.

Consider the sequence $(a_n)_{n \ge 0}$ which starts $7, 12, 17, 22, \ldots\text{.}$

What is the next term in the sequence?
Find a formula for the $n$th term of this sequence.

$a_n =$
Find the sum of the first 100 terms of the sequence: $\sum_{k=0}^{99}a_k\text{.}$

Solution.

$27\text{,}$ which is $22+5\text{.}$
The sequence is arithmetic, with $a_0 = 7$ and constant difference 5, so $a_n = 7 + 5n\text{.}$
$25450\text{.}$ We want $7 + 17 + \cdots + 502\text{.}$ Reverse and add to get 100 sums of 509, a total of 50900, which is twice the sum we are looking for.

3.

Consider the sum $8 + 19 + 30 + 41 + \cdots + 382\text{.}$

How many terms (summands) are in the sum?
Compute the sum using a technique discussed in this section.

Solution.

35.
$\displaystyle \frac{390 \cdot 34}{2} = 6825\text{.}$

4.

Consider the sequence $37, 48, 59, 70, \ldots, 11n + 4\text{.}$

How many terms are there in the sequence? Your answer will be in terms of $n\text{.}$
What is the second-to-last term?
Find the sum of all the terms in the sequence, in terms of $n\text{.}$

Solution.

${n-3+1}$ terms, since to get 37 using the formula $11n+4$ we must use $n=3\text{.}$
$11n - 7\text{,}$ which is 11 less than $11n+4$ (or plug in $n-1$ for $n$).
$\frac{(11n+41)*({n-3+1})}{2}\text{.}$ Reverse and add. Each sum gives the constant $11n+41$ and there are ${n-3+1}$ terms.

5.

Find $12 + 16 + 20 + 24+ \cdots + 1024$ using a technique from this section.

Solution.

$131572\text{.}$ If we take $a_0 = 12\text{,}$ the terms of the sum are an arithmetic sequence with closed formula $a_n = 12+4n\text{.}$ Then $1024 = a_{253}\text{,}$ for a total of 254 terms in the sum. Reverse and add to get 254 identical 1036 terms, which is twice the total we seek. $1036\cdot 254/2 = 131572$

6.

Find $2 + 10 + 50 + \cdots + 2\cdot 5^{20}\text{.}$

Solution.

$\frac{2\cdot 5^{21}-2}{4}\text{.}$ Let the sum be $S\text{,}$ and compute $5S - S = 4S\text{,}$ which causes terms except $2$ and $2\cdot 5^{21}$ to cancel. Then solve for $S\text{.}$

7.

Find $1 - \frac{2}{7} + \frac{4}{49} - \cdots +(-1)^{28} \frac{2^{28}}{7^{28}}\text{.}$

Solution.

$\frac{1 + (-1)^{29} \frac{2^{29}}{7^{29}}}{9/7}\text{.}$ Compute $S + \frac{2}{7}S\text{.}$

8.

Find $x$ and $y$ such that $64, x, y, 1$ is part of an arithmetic sequence.

$x =$ , $y =$

Then find $x$ and $y$ so that the sequence is part of a geometric sequence.

$x =$ , $y =$

(Warning: $x$ and $y$ might not be integers.)

Solution.

For arithmetic: $x = 43\text{,}$ $y = 22\text{.}$ For the arithmetic sequence, we know $64 + d = x\text{,}$ $x + d = y\text{,}$ and $y + d = 1\text{.}$ In other words, $64 + 3d = 1$ so $d = (-21)/3\text{.}$ For geometric: $x = {16}$ and $y = {4}\text{.}$ Similarly we can find the common ratio for the geometric sequence by solving $64\cdot r^3 = 1$ for $r\text{.}$

9.

Find $x$ and $y$ such that $3, x, y, 54$ is part of an arithmetic sequence.

$x =$ , $y =$

Then find $x$ and $y$ so that the sequence is part of a geometric sequence.

$x =$ , $y =$

(Warning: $x$ and $y$ might not be integers.)

Solution.

For arithmetic: $x = 20 \text{,}$ $y = 37 \text{.}$ We know that $3 + d = x\text{,}$ $x + d = y\text{,}$ and $y + d = 54\text{.}$ In other words, $3 + 3d = 54$ so $d = (51)/3\text{.}$

For geometric: $x = {7.86222}$ and $y = {20.6049}\text{.}$ We can find the common ratio for the geometric sequence by solving $3\cdot r^3 = 54$ for $r\text{.}$

Exercises Additional Exercises

1.

Is there a pair of integers $(a,b)$ such that $a, x_1, y_1, b$ is part of an arithmetic sequences and $a, x_2, y_2, b$ is part of a geometric sequence with $x_1, x_2, y_1, y_2$ all integers?

2.

Consider the sequence $2, 7, 15, 26, 40, 57, \ldots$ (with $a_0 = 2$). 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.

Solution.

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

3.

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:

$On the left, a 5x5 square to the right of a rectangle with base 2 and height 5, separated by a small gap. An arrow points to the right, where a rectangle of base 7 and height 5 is shown, including a dotted line representing where the square and triangle on the left were glued together.$

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.

4.

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 $t_1, t_2, \ldots\text{.}$
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.

5.

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:

$One square.$

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

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

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

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 $a_1 = 1\text{,}$ $a_2 = 5\text{,}$ 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.

6.

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 $4^7 - 3^7\text{.}$

Prev Top Next