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Question Number 104859 by qwertyu last updated on 24/Jul/20

Answered by Dwaipayan Shikari last updated on 24/Jul/20

$$S_n=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+....+\frac{n}{2^n}$$$$\frac{S_n}{2}=\frac{1}{2^2}+\frac{2}{2^3}+....+\frac{n-1}{2^n}+\frac{n}{2^{n+1}}$$$$\frac{S_n}{2}=(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....)-\frac{n}{2^{n+1}}$$$$S_n=2.\frac{1}{2}\left(\frac{1-2^{-n}}{\frac{1}{2}}\right)-\frac{n}{2^n}$$$$S_n=2(1-2^{-n})-\frac{n}{2^n}$$

Commented by Dwaipayan Shikari last updated on 24/Jul/20

$$Imisunderstoodthefirstone.Itiscorrected$$

Answered by abdomathmax last updated on 26/Jul/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{k}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{\mathrm{n}+1}-1}{\mathbf{x}-1}\mathbf{if}\mathbf{x}\ne 1$$$$\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{kx}}^{\mathrm{k}-1}=\frac{\mathrm{d}}{\mathrm{dx}}(...)=\frac{{\mathrm{nx}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}}+1}{{(\mathrm{x}-1)}^2}$$$$\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}{\mathrm{x}}^{\mathrm{k}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}\times ({\mathrm{nx}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}}+1)$$$$\mathrm{x}=\frac{1}{2}\mathrm{we}\mathrm{get}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{\mathrm{k}}{2^{\mathrm{k}}}=\frac{1}{2{(1-\frac{1}{2})}^2}\times (\frac{\mathrm{n}}{2^{\mathrm{n}+1}}-\frac{\mathrm{n}+1}{2^{\mathrm{n}}}+1)$$$$=\frac{2\mathrm{n}}{2^{\mathrm{n}+1}}-2\frac{\mathrm{n}+1}{2^{\mathrm{n}}}+2=\frac{\mathrm{n}}{2^{\mathrm{n}}}-\frac{\mathrm{n}+1}{2^{\mathrm{n}-1}}+2$$

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