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Question Number 99259 by DGmichael last updated on 19/Jun/20

Answered by abdomathmax last updated on 19/Jun/20

$${\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}{2}^{\mathrm{k}-1}\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{k}}\Rightarrow$$$${\mathrm{p}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{kx}}^{\mathrm{k}-1}\mathrm{we}\mathrm{have}\mathrm{p}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^{\mathrm{n}+1}-1}{\mathrm{x}-1}\Rightarrow$$$${\mathrm{p}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{(\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}}(\mathrm{x}-1)-({\mathrm{x}}^{\mathrm{n}+1}-1)}{{(\mathrm{x}-1)}^2}$$$$=\frac{(\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}+1}+1}{{(\mathrm{x}-1)}^2}$$$$=\frac{{\mathrm{nx}}^{\mathrm{n}+1}-(\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}}+1}{{(\mathrm{x}-1)}^2}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}={\mathrm{p}}^{{}^{\prime }}\left(2\right)=\frac{{\mathrm{n}2}^{\mathrm{n}+1}-(\mathrm{n}+1)2^{\mathrm{n}}+1}{1^2}$$$$=2\mathrm{n}{2}^{\mathrm{n}}-(\mathrm{n}+1)2^{\mathrm{n}}+1=(\mathrm{n}-1)2^{\mathrm{n}}+1$$

Commented by DGmichael last updated on 19/Jun/20

thanks sir !

Answered by mr W last updated on 19/Jun/20

$$S=\underset{k=1}{\sum ^n}k{2}^{k-1}$$$$2S=\underset{k=1}{\sum ^n}k{2}^k$$$$2S=\underset{k=2}{\sum ^{n+1}}(k-1)2^{k-1}$$$$2S=\underset{k=2}{\sum ^{n+1}}k{2}^{k-1}-\underset{k=2}{\sum ^{n+1}}{2}^{k-1}$$$$2S=\underset{k=1}{\sum ^n}k{2}^{k-1}+(n+1)2^n-1-\frac{2(2^n-1)}{2-1}$$$$2S=S+(n+1)2^n-1-2\times 2^n+2$$$$\Rightarrow S=(n-1)2^n+1$$

Commented by Ar Brandon last updated on 20/Jun/20

Cool, I erred��

Commented by DGmichael last updated on 20/Jun/20

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