# proof of sum

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• September 18th 2011, 05:13 PM
giygaskeptpraying
proof of sum
I'm sorry that the thread title isn't descriptive at all, but I didn't know what to name it.

I have to take this sum:
sum from 0 to log base 2 of &#40;n-1&#41; of &#40;n&#40;&#40;3&#47;2&#41;&#94;k&#41;&#41; - Wolfram|Alpha

and prove that for every positive integer n that is a power of 2, the sum is:
2(n^(log3/log2) - n)

can anyone help? thanks in advance
• September 20th 2011, 05:55 AM
CaptainBlack
Re: proof of sum
Quote:

Originally Posted by giygaskeptpraying
I'm sorry that the thread title isn't descriptive at all, but I didn't know what to name it.

I have to take this sum:
sum from 0 to log base 2 of &#40;n-1&#41; of &#40;n&#40;&#40;3&#47;2&#41;&#94;k&#41;&#41; - Wolfram|Alpha

and prove that for every positive integer n that is a power of 2, the sum is:
2(n^(log3/log2) - n)

can anyone help? thanks in advance

Your sum is:

$\sum_{k=0}^{\frac{\log(-1+n)}{\log(2)}} n.\left( \frac{3}{2}\right)^k$

1. Suppose $n=2^r$

2. The sum is not well defined as the upper limit of summation is not an integer, so I suppose it means:

$\left\lfloor \frac{\log(n-1)}{\log(2)}\right\rfloor=\lfloor \log_2(n-1)\rfloor = r-1$

3. The series is a finite geometric series.

CB