How to get the partial formula of $\sum_{k=1}^{n} k2^{-k}$? I know that the formula of a sequence of partial sum is from $k=i$, $\frac{q^i-q^{n+1}}{1-q}$.
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See the first answer here. – David Mitra Apr 26 '13 at 10:28
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$$f(x)=\sum_{k=0}^n x^k=\frac{1-x^{n+1}}{1-x}$$ and $$xf'(x)=\sum_{k=1}^{n}kx^k=x\frac{d}{dx}\left(\frac{1-x^{n+1}}{1-x}\right)$$
Take $x=\frac{1}{2}$ in the last equality.
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Start with: $$ \sum_{0 \le k \le n} z^k = \frac{1 - z^{n + 1}}{1 - z} $$ Then: $$ z \frac{d}{d z} \sum_{0 \le k \le z} z^k = \sum_{0 \le k \le n} k z^k $$ Compute the derivative of the left hand side by the alternative above and evaluate at $z = 1/2$.
vonbrand
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