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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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
1 Answers
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You are dealing with a geometric sum, for which we have $ \sum_{k=0}^{m-1} ar^k= a \, \frac{1-r^{m}}{1-r}. $ In your case $a=n$, $r=2^{-1}$ and $m=\log_2(n)$. Substituting this gives: $ n \, \frac{1-2^{-\log_2(n)}}{1-\frac{1}{2}}=n\, \frac{1-1/n}{\frac{1}{2}}=2n(1-1/n)=2(n-1) $
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0I did not see that (Oops!..You have already mentioned that) – 2012-05-10