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Question Number 102784 by bobhans last updated on 11/Jul/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n}{4^n}=?$$

Answered by Dwaipayan Shikari last updated on 11/Jul/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n}{4^n}=\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+\frac{5}{1024}+....=S_n$$$$\frac{S_n}{4}=\frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\frac{4}{1024}+......$$$$subtracting$$$$\frac{3S_n}{4}=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\frac{1}{1024}+......$$$$\frac{3S_n}{4}=\frac{1}{3}\Rightarrow S_n=\frac{4}{9}$$$$$$

Answered by bramlex last updated on 11/Jul/20

$$\mathrm{let}\mathrm{S}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}+1}{4^{\mathrm{n}}}\mathrm{\&}\mathrm{X}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}}{4^{\mathrm{n}}}$$$$\iff \mathrm{S}=\mathrm{X}+\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{4^{\mathrm{n}}}$$$$\mathrm{S}=\mathrm{X}+\frac{\frac{1}{4}}{1-\frac{1}{4}};\left(\mathrm{geometric}\mathrm{P}\right)$$$$\mathrm{S}=\mathrm{X}+\frac{1}{3}.\mathrm{in}\mathrm{the}\mathrm{other}\mathrm{hand}$$$$\mathrm{we}\mathrm{have}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}+1}{4^{\mathrm{n}}}=4\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}}{4^{\mathrm{n}}}-1$$$$\mathrm{S}=4\mathrm{X}-1=\mathrm{X}+\frac{1}{3}\Rightarrow 3\mathrm{X}=\frac{4}{3};\mathrm{X}=\frac{4}{9}$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}}{4^{\mathrm{n}}}=\frac{4}{9}\bullet$$$$$$

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

$$\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\mathrm{n}}{4^{\mathrm{n}}}\mathrm{let}\mathrm{w}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}\mathrm{with}\mid \mathrm{x}\mid <1\Rightarrow$$$${\mathrm{w}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}\Rightarrow {\mathrm{xw}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{\left(\frac{1}{4}\right)}^{\mathrm{n}}=\frac{1}{4}{\mathrm{w}}^{{}^{\prime }}\left(\frac{1}{4}\right)$$$$\mathrm{we}\mathrm{have}\mathrm{w}\left(\mathrm{x}\right)=\frac{1}{1-\mathrm{x}}\Rightarrow {\mathrm{w}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{{(1-\mathrm{x})}^2}\Rightarrow {\mathrm{xw}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\mathrm{n}}{4^{\mathrm{n}}}=\frac{1}{4{(1-\frac{1}{4})}^2}=\frac{1}{4}\times {\left(\frac{4}{3}\right)}^2=\frac{4}{9}$$

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