What is the value of the following sum: $$\sum\limits_{i=0}^{\ (\log_2(n))-1)}\frac{n}{2^i}$$ Can you show how to go about arriving at the answer?
What is the $\sum\limits_{i=0}^{\ (\log_2(n))-1)}\frac{n}{2^i}$?
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$\begingroup$
logarithms
exponential-sum
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0Do you mean to find $$\sum_{0}^{\log_{2}(n-1)}\left( \frac{n}{2^i}\right) $$ ? – 2012-05-10
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0Is that actually what you mean? Is $n$ from a certain set (say, the powers of 2?), in which case is your summation limit correct? – 2012-05-10
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1What happens if $\log_2(n-1)$ is not an integer? Do you mean $\lfloor \log_2(n-1) \rfloor$? – 2012-05-10
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2Formally, $$n\sum_{k=0}^m 2^{-k}=n\left(2-2^{-m}\right)$$ Replace $m$ with that expression you need... – 2012-05-10
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0Thanks for the comments, I edited the question to make it clearer I hope. – 2012-05-10
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0@J.M.: It is obviously the answer is given from your comment and the draks' one altogether. Is your statemnt supposed to be verified by induction? – 2012-05-10
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0@Babak: it's a truncated geometric series, no? There are standard methods for demonstrating that identity... – 2012-05-10
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0Maybe it helps, if @PaulCaheny would write out sum (without $\sum$) he's interested. If you restrict yourself to $n=2^m$ everthings fine. – 2012-05-10