Use the Formula for a Geometric Series

Finally, we will use formulas to find the terms of a geometric series.

Using the Formula for 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.

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

\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

Finding the First n Terms of a Geometric Series

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

S_{11} for the series 8 + -4 + 2 + …

\sum_{k=1}^{6} 3\cdot2^k


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

At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.


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
Creative Commons License This work is licensed under a Creative Commons Attribution 4.0 License.