Use the Formula for a Geometric Series

Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, $r$. We can write the sum of the first $n$ terms of a geometric series as

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

Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first $n$ terms of a geometric series. We will begin by multiplying both sides of the equation by $r$.

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

Next, we subtract this equation from the original equation.

$\frac{-r S_{R}=-\left(r a_{1}+r^{2} a_{1}+r^{3} a_{1}+\ldots+r^{R} a_{1}\right)}{(1-r) S_{R}=a_{1}-r^{2} a_{1}}$

Notice that when we subtract, all but the first term of the top equation and the last term of the bottorn equation cancel out. To obtain a formula for $S_{n}$, divide both sides by $(1-r)$.

$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r} \quad \Gamma \neq 1$

Formula for the Sum of the First $n$  Terms of a Geometric Series

A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first $n$ terms of a geometric sequence is represented as

$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r} \quad \Gamma \neq 1$

How To

Given a geometric series, find the sum of the first $n$ terms.

1. Identify $a_1,r$, and $n$.
   
2. Substitute values for $a_1,r$, and $n$ into the formula $S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}$.
   
3. Simplify to find $S_n$.
   

Example 4

Use the formula to find the indicated partial sum of each geometric series.

ⓐ $S_{11}$ for the series $8 + -4 + 2 + …$

Solution

ⓐ $a_1=8$, and we are given that $n=11$.

We can find $r$ by dividing the second term of the series by the first.

$r=\frac{-4}{8}=-\frac{1}{2}$

Substitute values for $a_{1}, r$, and $n$ into the formula and simplify.

$\begin{aligned} &S_{n}=\frac{a_{1}\left(1-r^{R}\right)}{1-r} \\ &S_{11}=\frac{8\left(1-\left(-\frac{1}{2}\right)^{11}\right)}{1-\left(-\frac{1}{2}\right)} \approx 5.336 \end{aligned}$

ⓑ Find $a_{1}$ by substituting $k=1$ into the given explicit formula.

$a_{1}=3 \cdot 2^{1}=6$

We can see from the given explicit formula that $r=2$. The upper limit of summation is 6 , so $n=6$. Substitute values for $a_{1}, \quad r$, and $n$ into the formula, and simplify.

$\begin{aligned} &S_{n}=\frac{a_{1}\left(1-r^{R}\right)}{1-r} \\ &S_{6}=\frac{6\left(1-2^{6}\right)}{1-2}=378 \end{aligned}$

Use the formula to find the indicated partial sum of each geometric series.

Try It #6

$S_{20}$ for the series $1,000 + 500 + 250 + …$

Try It #7

$\displaystyle\sum_{k=1}^{8} 3^k$

Example 5

Solving an Application Problem with a Geometric Series

Solution

The problem can be represented by a geometric series with $a_1=26,750; n=5$; and $r=1.016$. Substitute values for $a_1, r$, and $n$ into the formula and simplify to find the total amount earned at the end of 5 years.

$\begin{aligned}&S_{n}=\frac{a_{1}\left(1-r^{R}\right)}{1-r} \\&S_{5}=\frac{26,750\left(1-1.016^{6}\right)}{1-1.016} \approx 138,099.03\end{aligned}$

He will have earned a total of $138,099.03 by the end of 5 years.

Try It #8

At a new job, an employee’s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?

Source: Rice University, https://openstax.org/books/college-algebra/pages/9-4-series-and-their-notations
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