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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