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# Sums of Finite Geometric Series

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Finding the Sum of a Finite Geometric Series
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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? ### Guidance We have discussed in previous sections 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 $a_n=a_1r^{n-1}$ The sum of the first $n$ terms of a geometric sequence is: $S_n=a_1+a_1r+a_1r^2+a_1r^3+ \ldots +a_1r^{n-2}+a_1r^{n-1}$ Now, factor out $a_1$ to get $a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1})$ . If we isolate what is in the parenthesis and multiply this sum by $(1-r)$ as shown below we can simplify the sum: $&(1-r)S_n=(1-r)(1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \\&= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\&= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\&= (1-r)^n$ 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. $a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \frac{1-r}{1-r}=\frac{a_1(1-r^n)}{1-r}$ , which is the sum of a finite geometric series. So, $S_n=\frac{a_1(1-r^n)}{1-r}$ #### Example A 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, “Find $\sum\limits_{n=1}^{10} \frac{1}{32}(-2)^{n-1}$ .” Solution: Using the formula, $a_1=\frac{1}{32}$ , $r=-2$ , and $n=10$ . $S_{10}=\frac{\frac{1}{32}(1-(-2)^{10})}{1-(-2)} = \frac{\frac{1}{32}(1-1024)}{3} = -\frac{341}{32}$ We can also use the calculator as shown below. $sum(seq(1/32(-2)^{x-1}, x, 1, 10))=-\frac{341}{32}$ #### Example B Find the first term and the $n^{th}$ 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: $242 &= \frac{a_1(1-3^5)}{1-3} \\242 &= \frac{a_1(-242)}{-2} \\242 &= 121a_1 \\a_1 &= 2$ The first term is 2 and $a_n=2(3)^{n-1}$ . #### 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 $11^{th}$ 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 $1.08^2$ ):

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

$^*$ 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.

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

Intro Problem Revisit 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. ### Guided Practice 1. Evaluate $\sum\limits_{n=3}^8 2(-3)^{n-1}$ . 2. 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 $n^{th}$ 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 $49^{th}$ month).

1. Since we are asked to find the sum of the $3^{rd}$ through $8^{th}$ 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:

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

2. We can substitute what we know into the formula for the sum of a geometric series and solve for $a_1$ .

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

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

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

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

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 $n=49$ .

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

### Practice

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

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

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

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

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

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

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

1. 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? 2. 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.