7.4.1: Sums of Finite Geometric Series

Last updated
Save as PDF

Page ID: 14795

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Finding the Sum of a Finite Geometric Series

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 9^th month?

Sum of Finite Geometric Series

We have discussed how to use the calculator to find the sum of any series provided we know the n^th 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 $\ a_{n}=a_{1} r^{n-1}$.

The sum of the first $\ n$ terms of a geometric sequence is:

$\ S_{n}=a_{1}+a_{1} r+a_{1} r^{2}+a_{1} r^{3}+\ldots+a_{1} r^{n-2}+a_{1} r^{n-1}$.

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

$\ \begin{array}{l}
(1-r) S_{n}=(1-r)\left(1+r+r^{2}+r^{3}+\ldots+r^{n-2}+r^{n-1}\right) \\
=\left(1+r+r^{2}+r^{3}+\ldots+r^{n-2}+r^{n-1}-r-r^{2}-r^{3}-r^{4}-\ldots-r^{n-1}-r^{n}\right) \\
=\left(1+r+r^{2}+r^{3}+\ldots+r^{n-2}+r^{n-1}-r-r^{2}-r^{3}-r^{4}-\ldots-r^{n-1}-r^{n}\right) \\
=(1-r)^{n}
\end{array}$

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

$\ a_{1}\left(1+r^{2}+r^{3}+\ldots+r^{n-2}+r^{n-1}\right) \frac{1-r}{1-r}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}$, which is the sum of a finite geometric series.

So, $\ S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}$.

Let's find the sum of the first ten terms of the geometric sequence $\ a_{n}=\frac{1}{32}(-2)^{n-1}$. This could also be written as, "Let's find $\ \sum_{n=1}^{10} \frac{1}{32}(-2)^{n-1}$."

Using the formula, $\ a_{1}=\frac{1}{32}$, $\ r=-2$, and $\ n=10$.

$\ S_{10}=\frac{\frac{1}{32}\left(1-(-2)^{10}\right)}{1-(-2)}=\frac{\frac{1}{32}(1-1024)}{3}=-\frac{341}{32}$

We can also use the calculator as shown below.

$\ \operatorname{sum}\left(\operatorname{seq}\left(1 / 32(-2)^{x-1}, x, 1,10\right)\right)=-\frac{341}{32}$

Now, let's find the first term and the $\ n^{t h}$ term rule for a geometric series in which the sum of the first 5 terms is 242 and the common ratio is 3.

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

$\ \begin{aligned}
242 &=\frac{a_{1}\left(1-3^{5}\right)}{1-3} \\
242 &=\frac{a_{1}(-242)}{-2} \\
242 &=121 a_{1} \\
a_{1} &=2
\end{aligned}$

The first term is $\ 2$ and $\ a_{n}=2(3)^{n-1}$.

Finally, let's solve the following problem.

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 11^th year.

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 1.08²):

Sum Year 1: 1000

Sum Year 2: 1000 + 1000(1.08)

Sum Year 3: 1000 + 1000(1.08) + 1000(1.08)²

Sum Year 4: 1000 + 1000(1.08) + 1000(1.08)²+ 1000(1.08)³

$\ \quad\quad\quad\quad$⋮

Sum Year 11: 1000 + 1000(1.08) + 1000(1.08)²+ 1000(1.08)³+ … + 1000(1.08)⁹ + 1000(1.08)¹⁰

∗ There are 11 terms in this series because on the first day of the 11^th 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 $\ n=11$. Now we can calculate the sum using the formula and determine the value of the investment account at the start of the 11^th year.

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

Examples

Example 1

Earlier, you were asked to find how much money is in your account on the first day of the 9^th month.

Solution

There are 9 terms in this series because on the first day of the 9^th 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 n=9. Now we can calculate the sum using the formula and determine the value of the investment account at the start of the 9^th month.

$\ s_{9}=\frac{100\left(1-1.005^{9}\right)}{1-1.005}=918.22$

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

Example 2

Evaluate $\ \sum_{n=3}^{8} 2(-3)^{n-1}$

Solution

Since we are asked to find the sum of the $\ 3^{r d}$ through $\ 8^{t h}$ terms, we will consider $\ a_{3}$ as the first term. The third term is $\ a_{3}=2(-3)^{2}=2(9)=18$. Since we are starting with term three, we will be summing 6 terms, $\ a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}$, in total. We can use the rule for the sum of a geometric series now with $\ a_{1}=18$, $\ r=-3$ and $\ n=6$ to find the sum:

$\ \sum_{n=3}^{8} 2(-3)^{n-1}=\frac{18\left(1-(-3)^{6}\right)}{1-(-3)}=-3276$

Example 3

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

Solution

We can substitute what we know into the formula for the sum of a geometric series and solve for $\ a_{1}$.

$\ \begin{aligned}(l)
\frac{215}{8} &=\frac{a_{1}\left(1-\left(-\frac{1}{2}\right)^{7}\right)}{1-\left(-\frac{1}{2}\right)} \\
\frac{215}{8} &=a_{1}\left(\frac{43}{64}\right) \\
a_{1} &=\left(\frac{64}{43}\right)\left(\frac{215}{8}\right)=40
\end{aligned}$

The $\ n^{t h}$ term rule is $\ a_{n}=40\left(-\frac{1}{2}\right)^{n-1}$

Example 4

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 49^th month).

Solution

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

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

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 $\ a_{1}=50$, $\ r=1.005$ and $\ n=49$.

$\ S_{49}=\frac{50\left(1-(1.005)^{49}\right)}{(1-1.005)} \approx \$ 2768$

Review

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

$\ a_{n}=36\left(\frac{2}{3}\right)^{n-1}$
$\ a_{n}=9(-2)^{n-1}$
$\ a_{n}=5(-1)^{n-1}$
$\ a_{n}=\frac{8}{25}\left(\frac{5}{2}\right)^{n-1}$
$\ a_{n}=\frac{2}{3}\left(-\frac{3}{4}\right)^{n-1}$

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

$\ \sum_{n=1}^{4}(-1)\left(\frac{1}{2}\right)^{n-1}$
$\ \sum_{n=2}^{8}(128)\left(\frac{1}{4}\right)^{n-1}$
$\ \sum_{n=2}^{7} \frac{125}{64}\left(\frac{4}{5}\right)^{n-1}$
$\ \sum_{n=5}^{11} \frac{1}{32}(-2)^{n-1}$

Given the sum and the common ratio, find the $\ n^{t h}$ term rule for the series.

$\ \sum_{n=1}^{6} a_{n}=-63$ and $\ r=-2$
$\ \sum_{n=1}^{4} a_{n}=671$ and $\ r=\frac{5}{6}$
$\ \sum_{n=1}^{5} a_{n}=122$ and $\ r=-3$
$\ \sum_{n=2}^{7} a_{n}=-\frac{63}{2}$ and $\ r=-\frac{1}{2}$

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 5^th birthday. They continue to make this birthday deposit each year until making the final deposit on her 18^th 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 6^th year. Round your answer to the nearest $100.

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 11.10.

Vocabulary

Term	Definition
induction	Induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers.
series	A series is the sum of the terms of a sequence.