You are saving for summer camp. You deposit $100 on the first of each month into your savings account. The account grows at a rate of 0.5% per month. How much money is in your account on the first day on the \begin{align*}9^{th}\end{align*} month?

### Guidance

We have discussed in previous sections how to use the calculator to find the sum of any series provided we know the @$\begin{align*}n^{th}\end{align*}@$ term rule. For a geometric series, however, there is a specific rule that can be used to find the sum algebraically. Let’s look at a finite geometric sequence and derive this rule.

Given @$\begin{align*}a_n=a_1r^{n-1}\end{align*}@$

The sum of the first @$\begin{align*}n\end{align*}@$ terms of a geometric sequence is: @$\begin{align*}S_n=a_1+a_1r+a_1r^2+a_1r^3+ \ldots +a_1r^{n-2}+a_1r^{n-1}\end{align*}@$

Now, factor out @$\begin{align*}a_1\end{align*}@$ to get @$\begin{align*}a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1})\end{align*}@$ . If we isolate what is in the parenthesis and multiply this sum by @$\begin{align*}(1-r)\end{align*}@$ as shown below we can simplify the sum:

@$$\begin{align*}&(1-r)S_n=(1-r)(1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \\ &= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= (1-r)^n\end{align*}@$$

By multiplying the sum by @$\begin{align*}1-r\end{align*}@$ we were able to cancel out all of the middle terms. However, we have changed the sum by a factor of @$\begin{align*}1-r\end{align*}@$ , so what we really need to do is multiply our sum by @$\begin{align*}\frac{1-r}{1-r}\end{align*}@$ , or 1.

@$\begin{align*}a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \frac{1-r}{1-r}=\frac{a_1(1-r^n)}{1-r}\end{align*}@$ , which is the sum of a finite geometric series.

So, @$\begin{align*}S_n=\frac{a_1(1-r^n)}{1-r}\end{align*}@$

#### Example A

Find the sum of the first ten terms of the geometric sequence @$\begin{align*}a_n=\frac{1}{32}(-2)^{n-1}\end{align*}@$ . This could also be written as, “Find @$\begin{align*}\sum\limits_{n=1}^{10} \frac{1}{32}(-2)^{n-1}\end{align*}@$ .”

**
Solution:
**
Using the formula,
@$\begin{align*}a_1=\frac{1}{32}\end{align*}@$
,
@$\begin{align*}r=-2\end{align*}@$
, and
@$\begin{align*}n=10\end{align*}@$
.

@$$\begin{align*}S_{10}=\frac{\frac{1}{32}(1-(-2)^{10})}{1-(-2)} = \frac{\frac{1}{32}(1-1024)}{3} = -\frac{341}{32}\end{align*}@$$

We can also use the calculator as shown below.

@$$\begin{align*}sum(seq(1/32(-2)^{x-1}, x, 1, 10))=-\frac{341}{32}\end{align*}@$$

#### Example B

Find the first term and the @$\begin{align*}n^{th}\end{align*}@$ term rule for a geometric series in which the sum of the first 5 terms is 242 and the common ratio is 3.

**
Solution:
**
Plug in what we know to the formula for the sum and solve for the first term:

@$$\begin{align*}242 &= \frac{a_1(1-3^5)}{1-3} \\ 242 &= \frac{a_1(-242)}{-2} \\ 242 &= 121a_1 \\ a_1 &= 2\end{align*}@$$

The first term is 2 and @$\begin{align*}a_n=2(3)^{n-1}\end{align*}@$ .

#### Example C

Charlie deposits $1000 on the first of each year into his investment account. The account grows at a rate of 8% per year. How much money is in the account on the first day on the @$\begin{align*}11^{th}\end{align*}@$ year.

**
Solution:
**
First, consider what is happening here on the first day of each year. On the first day of the first year, $1000 is deposited. On the first day of the second year $1000 is deposited and the previously deposited $1000 earns 8% interest or grows by a factor of 1.08 (108%). On the first day of the third year another $1000 is deposited, the previous year’s deposit earns 8% interest and the original deposit earns 8% interest for two years (we multiply by
@$\begin{align*}1.08^2\end{align*}@$
):

@$$\begin{align*}&\text{Sum Year} \ 1: 1000\\ &\text{Sum Year} \ 2:1000 + 1000(1.08)\\ &\text{Sum Year} \ 3: 1000 + 1000(1.08) + 1000(1.08)^2\\ &\text{Sum Year} \ 4: 1000 + 1000(1.08) + 1000(1.08)^2 + 1000(1.08)^3\\ & \qquad \qquad \qquad \qquad \qquad \vdots\\ &\text{Sum Year} \ 11: 1000 + 1000(1.08) + 1000(1.08)^2 + 1000(1.08)^3 + \ldots + 1000(1.08)^9 + 1000(1.08)^{10}\end{align*}@$$

@$\begin{align*}^*\end{align*}@$ There are 11 terms in this series because on the first day of the @$\begin{align*}11^{th}\end{align*}@$ year we make our final deposit and the original deposit earns interest for 10 years.

This series is geometric. The first term is 1000, the common ratio is 1.08 and @$\begin{align*}n=11\end{align*}@$ . Now we can calculate the sum using the formula and determine the value of the investment account at the start of the @$\begin{align*}11^{th}\end{align*}@$ year.

@$$\begin{align*}s_{11}=\frac{1000 \left(1-1.08^{11}\right)}{1-1.08}=16645.48746 \approx \$16,645.49\end{align*}@$$

**
Intro Problem Revisit
**
There are 9 terms in this series because on the first day of the
@$\begin{align*}9^{th}\end{align*}@$
month you make your final deposit and the original deposit earns interest for 8 months.

This series is geometric. The first term is 100, the common ratio is 1.005 and @$\begin{align*}n=9\end{align*}@$ . Now we can calculate the sum using the formula and determine the value of the investment account at the start of the @$\begin{align*}9^{th}\end{align*}@$ month.

@$$\begin{align*}s_{9}=\frac{100 \left(1-1.005^{9}\right)}{1-1.005}= 918.22\end{align*}@$$

Therefore there is $918.22 in the account at the beginning of the ninth month.

### Guided Practice

1. Evaluate @$\begin{align*}\sum\limits_{n=3}^8 2(-3)^{n-1}\end{align*}@$ .

2. If the sum of the first seven terms in a geometric series is @$\begin{align*}\frac{215}{8}\end{align*}@$ and @$\begin{align*}r=-\frac{1}{2}\end{align*}@$ , find the first term and the @$\begin{align*}n^{th}\end{align*}@$ term rule.

3. Sam deposits $50 on the first of each month into an account which earns 0.5% interest each month. To the nearest dollar, how much is in the account right after Sam makes his last deposit on the first day of the fifth year (the @$\begin{align*}49^{th}\end{align*}@$ month).

#### Answers

1. Since we are asked to find the sum of the @$\begin{align*}3^{rd}\end{align*}@$ through @$\begin{align*}8^{th}\end{align*}@$ terms, we will consider @$\begin{align*}a_3\end{align*}@$ as the first term. The third term is @$\begin{align*}a_3=2(-3)^2=2(9)=18\end{align*}@$ . Since we are starting with term three, we will be summing 6 terms, @$\begin{align*}a_3+a_4+a_5+a_6+a_7+a_8\end{align*}@$ , in total. We can use the rule for the sum of a geometric series now with @$\begin{align*}a_1=18\end{align*}@$ , @$\begin{align*}r=-3\end{align*}@$ and @$\begin{align*}n=6\end{align*}@$ to find the sum:

@$$\begin{align*}\sum_{n=3}^8 2(-3)^{n-1}=\frac{18(1-(-3)^6)}{1-(-3)}=-3276\end{align*}@$$

2. We can substitute what we know into the formula for the sum of a geometric series and solve for @$\begin{align*}a_1\end{align*}@$ .

@$$\begin{align*}\frac{215}{8} &= \frac{a_1 \left(1-\left(-\frac{1}{2}\right)^7\right)}{1- \left(-\frac{1}{2}\right)} \\ \frac{215}{8} &= a_1 \left(\frac{43}{64}\right) \\ a_1 &= \left(\frac{64}{43}\right)\left(\frac{215}{8}\right)=40\end{align*}@$$

The @$\begin{align*}n^{th}\end{align*}@$ term rule is @$\begin{align*}a_n=40 \left(-\frac{1}{2}\right)^{n-1}\end{align*}@$

3. The deposits that Sam make and the interest earned on each deposit generate a geometric series,

@$$\begin{align*}& S_{49}=50+50(1.005)^1+50(1.005)^2+50(1.005)^3+ \ldots +50(1.005)^{47}+50(1.005)^{48}, \\ & \qquad \ \ \uparrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \uparrow \\ & \quad \text{last deposit} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{first deposit}\end{align*}@$$

Note that the first deposit earns interest for 48 months and the final deposit does not earn any interest. Now we can find the sum using @$\begin{align*}a_1=50\end{align*}@$ , @$\begin{align*}r=1.005\end{align*}@$ and @$\begin{align*}n=49\end{align*}@$ .

@$$\begin{align*}S_{49}=\frac{50(1-(1.005)^{49})}{(1-1.005)} \approx \$2768\end{align*}@$$

### Explore More

Use the formula for the sum of a geometric series to find the sum of the first five terms in each series.

- @$\begin{align*}a_n=36 \left(\frac{2}{3}\right)^{n-1}\end{align*}@$
- @$\begin{align*}a_n=9(-2)^{n-1}\end{align*}@$
- @$\begin{align*}a_n=5(-1)^{n-1}\end{align*}@$
- @$\begin{align*}a_n=\frac{8}{25} \left(\frac{5}{2}\right)^{n-1}\end{align*}@$
- @$\begin{align*}a_n=\frac{2}{3} \left(-\frac{3}{4}\right)^{n-1}\end{align*}@$

Find the indicated sums using the formula and then check your answers with the calculator.

- @$\begin{align*}\sum\limits_{n=1}^4(-1) \left(\frac{1}{2}\right)^{n-1}\end{align*}@$
- @$\begin{align*}\sum\limits_{n=2}^8(128) \left(\frac{1}{4}\right)^{n-1}\end{align*}@$
- @$\begin{align*}\sum\limits_{n=2}^7 \frac{125}{64}\left(\frac{4}{5}\right)^{n-1}\end{align*}@$
- @$\begin{align*}\sum\limits_{n=5}^{11} \frac{1}{32}(-2)^{n-1}\end{align*}@$

Given the sum and the common ratio, find the @$\begin{align*}n^{th}\end{align*}@$ term rule for the series.

- @$\begin{align*}\sum\limits_{n=1}^6 a_n=-63\end{align*}@$ and @$\begin{align*}r=-2\end{align*}@$
- @$\begin{align*}\sum\limits_{n=1}^4 a_n=671\end{align*}@$ and @$\begin{align*}r=\frac{5}{6}\end{align*}@$
- @$\begin{align*}\sum\limits_{n=1}^5 a_n=122\end{align*}@$ and @$\begin{align*}r=-3\end{align*}@$
- @$\begin{align*}\sum\limits_{n=2}^7 a_n=-\frac{63}{2}\end{align*}@$ and @$\begin{align*}r=-\frac{1}{2}\end{align*}@$

Solve the following word problems using the formula for the sum of a geometric series.

- Sapna’s grandparents deposit $1200 into a college savings account on her @$\begin{align*}5^{th}\end{align*}@$ birthday. They continue to make this birthday deposit each year until making the final deposit on her @$\begin{align*}18^{th}\end{align*}@$ birthday. If the account earns 5% interest annually, how much is there after the final deposit?
- Jeremy wants to have save $10,000 in five years. If he makes annual deposits on the first of each year and the account earns 4.5% interest annually, how much should he deposit each year in order to have $10,000 in the account after the final deposit on the first of the @$\begin{align*}6^{th}\end{align*}@$ year. Round your answer to the nearest $100.