SOLUTION: what is the sum of the geometric series, rounded to the nearest whole number? 16E(n-1) 6(1/4)^n

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Question 1118291: what is the sum of the geometric series, rounded to the nearest whole number?
16E(n-1) 6(1/4)^n
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Describe the series in words, instead of making us try to decipher your nonstandard symbols.

The only possible logical interpretation I can see is the sum as n goes from 1 to 16 of 6%281%2F4%29%5En.

The infinite sum is (first term) divided by (1 minus the common ratio). The first term is 6(1/4) = 3/2; the common ratio is 1/4:

S+=+%283%2F2%29%2F%281-1%2F4%29+=+%283%2F2%29%2F%283%2F4%29+=+%283%2F2%29%2A%284%2F3%29+=+2

The terms after n=16 are tiny and will certainly bring the sum down only a tiny bit.

So to the nearest whole number the sum of the finite geometric series is 2.