SOLUTION: Evaluate the sum:
4/sigma/n=1 (2/3)^n-1
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Question 1084210: Evaluate the sum:
4/sigma/n=1 (2/3)^n-1
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'm assuming the summation you want to do is
^{n-1})
If n = 1, then (2/3)^(n-1) = (2/3)^(1-1) = (2/3)^0 = 1
If n = 2, then (2/3)^(n-1) = (2/3)^(2-1) = (2/3)^1 = 2/3
If n = 3, then (2/3)^(n-1) = (2/3)^(3-1) = (2/3)^2 = 4/9
If n = 4, then (2/3)^(n-1) = (2/3)^(4-1) = (2/3)^3 = 8/27
Therefore, summing from n = 1 to n = 4 yields
^{n-1} \ \ = (1)+(2/3)+(4/9)+(8/27))
^{n-1} \ \ = (27/27)+(18/27)+(12/27)+(8/27))
^{n-1} \ \ = (27+18+12+8)/27)
^{n-1} \ \ = 65/27)
The answer would be the fraction 65/27.
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