**Sum Formulae for finite Geometric Series**

If *a*_{1}, *a*_{2}, *a*_{3}, ⋯, *a _{n}* is a finite geometric sequence, then the corresponding series

*a*

_{1}+

*a*

_{2}+

*a*

_{3}+⋯+

*a*is called a geometric series. As with arithmetic series, we can derive two simple and very useful formulas for the sum of a geometric series. Let

_{n}*r*be the common ratio of the geometric sequence

*a*

_{1},

*a*

_{2},

*a*

_{3}, ⋯,

*a*and let

_{n}*S*denote the sum of the series

_{n}*a*

_{1}+

*a*

_{2}+

*a*

_{3}+⋯+

*a*. Then

_{n}Multiply both sides of this equation by

*r*to obtain

Now subtract the left side of the second equation from the left side of the first, and the right side of the second equation from the right side of the first to obtain

*S*–

_{n}*r*⋅

*S*=

_{n}*a*

_{1}–

*a*

_{1}⋅

*r*

^{n}*S*(1-

_{n}*r*)=

*a*

_{1}–

*a*

_{1}⋅

*r*

^{n}Thus, solving for

*S*we obtain the following formula for the sum of a geometric series:

_{n}Theorem A:

Sum of a Geometric Series—first Form

Since *a _{n}*=

*a*

_{1}

*r*

^{(n-1)}, or

*a*

_{n}*r*=

*a*

_{1}⋅

*r*, the sum formula also can be written in the following form:

^{n}Theorem B:

Sum of a Geometric Series—Second Form

If *r*=1, then

*S*=

_{n}*a*

_{1}+

*a*

_{1}(1)+

*a*

_{1}(1

^{2})+⋯+

*a*

_{1}(1

^{(n-1)})=

*n*⋅

*a*

_{1}

How can you find the sum of a geometric series if you know the first and last terms and the common ratio, but not the number of terms? Remember the formula for the *n*th term of a geometric sequence or series, *a _{n}*=

*a*

_{1}⋅

*r*

^{(n-1)}. You can use this formula to find an expression involving

*r*.

^{n}Formula for

*n*th term,

*a*=

_{n}*a*

_{1}⋅

*r*

^{(n-1)}.

Multiply each side by

*r*.

*a*⋅

_{n}*r*=

*a*

_{1}⋅

*r*

^{(n-1)}⋅

*r*

*a*⋅

_{n}*r*=

*a*

_{1}⋅

*r*

^{n}Now substitute

*a*⋅

_{n}*r*for

*a*

_{1}⋅

*r*in the formula for the sum of a geometric series.

^{n}The result is

**an Alternate Formula For Sum of Finite Geometric Series — Second Form**

As is the case with arithmetic series, it is often desirable to find a general expression for the

*n*th partial sum of a geometric series.

Let us start with an example:

find the partial sum for the first 20 terms of the series

We express

*S*

_{20}in two different ways and subtract them:

This reasoning can be extended to any geometric series in order to develop a formula for the

*n*th partial sum

*S*.

_{n}Let {*a _{n}*} be a geometric sequence with first term

*a*

_{1}and a common ratio

*r*≠1. We can construct the series in two ways as before and using the definition of the geometric sequence, i.e.

*a*=

_{n}*a*

_{(n-1)}⋅

*r*, then

Now, we subtract the first and last expressions to get

This expression, however, requires that

*r*,

*a*

_{1}, as well as

*a*be known in order to find the sum.

_{n}Example 1: **use the Alternate Formula for a Sum**

find the sum of a geometric series for which *a*_{1}=15,625, *a _{n}*=-5, and

*r*=-⅕. Since you do not know the value of

*n*, use the formula derived above.

Ex2. Given a geometric sequence with

*a*=729 and 7th term 64, determine

*S*

_{7}.

Solution:

*a*=729,

*a*

_{7}=64.

Let

*r*be the common ratio of the geometric sequence. It is known that,

*a*=

_{n}*a*⋅

*r*

^{(n-1)},

Example 3: **find the sum of each geometric series described**.

Ex3a. first *n* terms of *a*_{1}=4, *a _{n}*=2000,

*r*=-3.

Solution:

Use

**Second Form**for the

*n*th partial sum of a geometric series.

Ex3b. first

*n*terms of

*a*

_{1}=-36,

*a*=972,

_{n}*r*=7.

Solution:

Use

**Second Form**for the

*n*th partial sum of a geometric series.

Ex3c. first *n* terms of *a*_{1}=-8, *a _{n}*=-256,

*r*=2

Solution:

Use

**Second Form**for the

*n*th partial sum of a geometric series.

Ex3d. first *n* terms of *a*_{1}=5, *a _{n}*=1,310,720,

*r*=4.

Solution:

Use

**Second Form**for the

*n*th partial sum of a geometric series.

Ex3e. first *n* terms of *a*_{1}=3, *a _{n}*=46,875,

*r*=-5.

Solution:

Use

**Second Form**for the

*n*th partial sum of a geometric series.

We now turn our attention to the sum of every geometric sequence.

To derive the formula for the geometric sum, We start with a geometric sequence *a _{k}*=

*a*⋅

*r*,

^{k}*k*≥1, and let

*S*once again denote the sum of the first

*n*terms. Comparing

*S*and

*r⋅S*, we get

Subtracting the second equation from the first forces all of the terms except

*a*and

*a*

*r*to cancel out and We get

^{n}*S-r⋅S=a-a*⋅

*r*. Factoring, we get

^{n}*S*(1-

*r*)=

*a*(1-

*r*). Assuming

^{n}*r*≠1, we can divide both sides by the quantity (1-

*r*) to obtain

If we distribute

*a*through the numerator, we get

*a*–

*a*

*r*=

^{n}*a*

_{1}–

*a*

_{(n+1)}which yields the formula

In the case when

*r*=1, we get the formula

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