$$1 + 2/2 + 3/4 + \cdots + n/2^{n-1}$$
How would find the closed-form expression and also the sum up to 20? I'm not really getting why or the logic behind using derivatives to arrive at an answer.
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This should give you plenty of ideas. – David Mitra Feb 12 '13 at 12:58
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That question is not identical; answers are not asked to give the formula for a finite sum. I do see that within Eric Naslund's answer is a formula similar to mine for a finite sum; however, that answer does not use derivatives, which is one of the points of this question. In any case, I don't think this is really a duplicate, but I won't act unilaterally on it. – robjohn Feb 12 '13 at 15:41
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The standard geometric series is $$ \sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}\tag{1} $$ Taking the derivative yields $$ \begin{align} \sum_{k=0}^{n}kx^{k-1} &=\frac{-(n+1)x^{n}(1-x)-(1-x^{n+1})(-1)}{(1-x)^2}\\ &=\frac{1-(n+1)x^{n}+nx^{n+1}}{(1-x)^2}\tag{2} \end{align} $$ Use $x=\frac12$ in $(2)$.
robjohn
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@RickDecker: which is what I wrote: Use $\color{#C00000}{x=\frac12}$ in $(2)$. – robjohn Feb 12 '13 at 14:28
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Oops. I read that as $x=\frac{1}{2}\ln(2)$. Guess I should get my eyes checked. – Rick Decker Feb 12 '13 at 15:25
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This is a arithmetico-geometric sequence and the computation of its sum is very standard : http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence
