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Question Number 98096 by Algoritm last updated on 11/Jun/20

Answered by mathmax by abdo last updated on 11/Jun/20

$$\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{3\mathrm{n}-1}{2^{\mathrm{n}-1}}{=}_{\mathrm{n}-1=\mathrm{p}}\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{3\mathrm{p}+2}{2^{\mathrm{p}}}$$$$\mathrm{so}\mathrm{S}=3\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{\mathrm{n}}{2^{\mathrm{n}}}+2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2^{\mathrm{n}}}=\frac{1}{1-\frac{1}{2}}=2$$$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}\mathrm{with}\mid \mathrm{x}\mid <1\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}=\frac{1}{1-\mathrm{x}}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}=\frac{1}{{(1-\mathrm{x})}^2}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\mathrm{n}{\left(\frac{1}{2}\right)}^{\mathrm{n}}=\frac{1}{2{(1-\frac{1}{2})}^2}=\frac{1}{2\times \frac{1}{4}}=2\Rightarrow \mathrm{S}=3\times 2)+(2\times 2$$$$\mathrm{S}=10$$

Commented by Algoritm last updated on 11/Jun/20

$$\mathrm{an}\mathrm{elementary}\mathrm{solution}\mathrm{was}\mathrm{needed}$$$$2\mathrm{S}-\mathrm{S}=?$$

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