Write the given series in the form of a geometric series. g

If a sequence is geometric there are ways to find the sum of the first nn terms, denoted SnSn, without actually adding all of the terms.

To find the sum of the first SnSn terms of a geometric sequence use the formula 
Sn=a1(1−rn)1−r,r≠1Sn=a1(1−rn)1−r,r≠1,
where nn is the number of terms, a1a1 is the first term and rr is the common ratio.

Example 1:

Find the sum of the first 8 terms of the geometric series if a1=1a1=1 and r=2r=2.

S8=1(1−28)1−2=255S8=1(1−28)1−2=255

Example 2:

Find S10S10 of the geometric sequence 24,12,6,⋯24,12,6,⋯.

First, find rr. 

r=r2r1=1224=12r=r2r1=1224=12

Now, find the sum:

S10=24(1−(12)10)1−12=306964S10=24(1−(12)10)1−12=306964

Example 3:

Evaluate.

∑n=1103(−2)n−1∑n=1103(−2)n−1

(You are finding S10S10 for the series 3−6+12−24+⋯3−6+12−24+⋯, whose common ratio is −2−2.)

Sn=a1(1−rn)1−rS10=3[1−(−2)10]1−(−2)=3(1−1024)3=−1023Sn=a1(1−rn)1−rS10=3[1−(−2)10]1−(−2)=3(1−1024)3=−1023  

In order for an infinite geometric series to have a sum, the common ratio rr must be between −1−1 and 11.  Then as nn increases, rnrn gets closer and closer to 00.  To find the sum of an infinite geometric series having ratios with an absolute value less than one, use the formula, S=a11−rS=a11−r, where a1a1 is the first term and rr is the common ratio.

Example 4:

Find the sum of the infinite geometric sequence
27,18,12,8,⋯27,18,12,8,⋯.

First find rr: 

r=a2a1=1827=23r=a2a1=1827=23

Then find the sum:

S=a11−rS=a11−r

S=271−23=81S=271−23=81

Example 5:

Find the sum of the infinite geometric sequence
8,12,18,27,⋯8,12,18,27,⋯ if it exists.

First find rr: 

r=a2a1=128=32r=a2a1=128=32

Since r=32r=32 is not less than one the series has no sum.

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step-by-step explanation: