SOLUTION: Evaluate the sum: 4/sigma/n=1 (2/3)^n-1

Algebra ->  Sequences-and-series -> SOLUTION: Evaluate the sum: 4/sigma/n=1 (2/3)^n-1      Log On


   

Question 1084210: Evaluate the sum:
4/sigma/n=1 (2/3)^n-1
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

I'm assuming the summation you want to do is


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









The answer would be the fraction 65/27.