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December 19, 2014

December 19, 2014

Posted by **relle** on Tuesday, April 24, 2007 at 6:39pm.

the way I am reading the question you have i is going from 2 to n

so there will be n-1 terms

using i = 2,3,4,..,n

I get 2^(2n-2) + 2^(2n-2) + ...

Sum of n terms of a geometric series is

a(1 - r^n)/(1-r)

so a= 2^(2n-2), r = 2^(-1) and the number of terms is n-1

and the sum is

2^(2n-2)[1 - 2^(-1)] / 1-2^(-1))

.

.

.

=2^(n-1)(2^n - 1)

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