SOLUTION: Find the sum of this infinite geometric series: {{{ 6-(6/2)+(6/4)-(6/8)+ . . . }}} --This is new to me. What is involved in solving this type of problem? Any directions/tips are

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Question 1093240: Find the sum of this infinite geometric series: +6-%286%2F2%29%2B%286%2F4%29-%286%2F8%29%2B+.+.+.+
--This is new to me. What is involved in solving this type of problem? Any directions/tips are appreciated :)
Found 3 solutions by josgarithmetic, greenestamps, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
common ratio, -1%2F2

If follow the formula, with the factor 6 on all the terms,

6%281%2F%281-%28-1%2F2%29%29%29
6%281%2F%281%2B1%2F2%29%29
6%281%2F%283%2F2%29%29
6%282%2F3%29
%286%2A2%29%2F3
2%2A2
highlight%284%29

Answer by greenestamps(13214) About Me  (Show Source):
You can put this solution on YOUR website!

In an infinite geometric series, if the common ratio r is between -1 and +1, then the series has a finite sum, given by the formula

S+=+a%2F%281-r%29
where a is the first term and r is the common ratio.

For your series, the first term is 6 and the common ratio is -1/2. So the sum is
6%2F%281%2B1%2F2%29+=+6%2F%283%2F2%29+=+6%2A%282%2F3%29+=+4

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!

Find the sum of this infinite geometric series: +6-%286%2F2%29%2B%286%2F4%29-%286%2F8%29%2B+.+.+.+
--This is new to me. What is involved in solving this type of problem? Any directions/tips are appreciated :) 
Sum of an infinite geometric series, or matrix%281%2C3%2C+S%5Binfinity%5D%2C+%22=%22%2C+a%5B1%5D%2F%281+-+r%29%29, where:

a%5B1%5D = 1st term (6, in this case)

r  = Common Ratio (-+1%2F2, in this case)

matrix%281%2C3%2C+S%5Binfinity%5D%2C+%22=%22%2C+6%2F%281+-+%28-+1%2F2%29%29%29 ------- Substituting matrix%281%2C7%2C+6%2C+for%2C+a%5B1%5D%2C+and%2C+-+1%2F2%2C+for%2C+r%29