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Question Number 141791 by Chhing last updated on 23/May/21

$$\mathrm{Calculate}$$$$\underset{\mathrm{n}=1}{\sum ^{+\mathrm{\infty }}}\frac{{\mathrm{n}}^2}{3^{\mathrm{n}}}$$

Answered by mr W last updated on 23/May/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^n=x+x^2+x^3+...=\frac{x}{1-x}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^{n-1}=\frac{1}{1-x}+\frac{x}{{(1-x)}^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^n=\frac{x}{1-x}+\frac{x^2}{{(1-x)}^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^2{x}^{n-1}=\frac{1}{1-x}+\frac{x}{{(1-x)}^2}+\frac{2x}{{(1-x)}^2}+\frac{2x^2}{{(1-x)}^3}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^2{x}^n=\frac{x}{1-x}+\frac{3x^2}{{(1-x)}^2}+\frac{2x^3}{{(1-x)}^3}$$$$withx=\frac{1}{3}:$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2}{3^n}=\frac{1}{2}+\frac{3}{4}+\frac{2}{8}=\frac{3}{2}$$

Answered by qaz last updated on 23/May/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2}{3^n}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(n+1)}^2}{3^{n+1}}=\frac{1}{3}{(xD+1)}^2{\mid }_{x=1/3}\frac{1}{1-x}$$$$=\frac{1}{3}{[{\left(xD\right)}^2+2xD+1]}_{x=1/3}\frac{1}{1-x}$$$$=\frac{1}{3}(x^2{D}^2+3xD+1){\mid }_{x=1/3}\frac{1}{1-x}$$$$=\frac{1}{3}{[\frac{2x^2}{{(1-x)}^3}+\frac{3x}{{(1-x)}^2}+\frac{1}{1-x}]}_{x=1/3}$$$$=\frac{1}{3}(\frac{3}{4}+\frac{9}{4}+\frac{3}{2})$$$$=\frac{3}{2}$$

Answered by mathmax by abdo last updated on 23/May/21

$$\mathrm{let}\mathrm{s}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}\mathrm{with}[\mid \mathrm{x}\mid <1\Rightarrow {\mathrm{s}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}\Rightarrow$$$${\mathrm{xs}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}\Rightarrow {\mathrm{s}}^{{}^{\prime }}\left(\mathrm{x}\right)+{\mathrm{xs}}^{\left(2\right)}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{n}}^2{\mathrm{x}}^{\mathrm{n}-1}\Rightarrow$$$${\mathrm{xs}}^{{}^{\prime }}\left(\mathrm{x}\right)+{\mathrm{x}}^2{\mathrm{s}}^{\left(2\right)}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{n}}^2{\mathrm{x}}^{\mathrm{n}}$$$$\mathrm{x}=\frac{1}{3}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{n}}^2}{3^{\mathrm{n}}}=\frac{1}{3}{\mathrm{s}}^{{}^{\prime }}\left(\frac{1}{3}\right)+\frac{1}{9}{\mathrm{s}}^{{}^{″}}\left(\frac{1}{3}\right)\mathrm{we}\mathrm{have}$$$$\mathrm{s}\left(\mathrm{x}\right)=\frac{1}{1-\mathrm{x}}\Rightarrow {\mathrm{s}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{{(1-\mathrm{x})}^2}\mathrm{and}{\mathrm{s}}^{″}\left(\mathrm{x}\right)=-\frac{2(1-\mathrm{x})(-1)}{{(1-\mathrm{x})}^4}$$$$=\frac{2}{{(1-\mathrm{x})}^3}\mathrm{so}{\mathrm{s}}^{{}^{\prime }}\left(\frac{1}{3}\right)=\frac{1}{{\left(\frac{2}{3}\right)}^2}=\frac{9}{4}\mathrm{and}{\mathrm{s}}^{\left(2\right)}\left(\frac{1}{3}\right)=2\times \frac{27}{8}=\frac{27}{4}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{n}}^2}{3^{\mathrm{n}}}=\frac{1}{3}.\frac{9}{4}+\frac{1}{9}.\frac{27}{4}=\frac{3}{4}+\frac{3}{4}=\frac{6}{4}=\frac{3}{2}$$

Answered by qaz last updated on 23/May/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(n+1)}^2}{3^{n+1}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(n+1)}^2}{\mathrm{\Gamma }(n+1)}{\int }_0^{\mathrm{\infty }}{u}^n{e}^{-3u}du$$$$={\int }_0^{\mathrm{\infty }}{e}^{-3u}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(n+1)}^2}{n!}{u}^{n}du$$$$={\int }_0^{\mathrm{\infty }}{e}^{-3u}{(xD+1)}^2{\mid }_{x=u}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}du$$$$={\int }_0^{\mathrm{\infty }}{e}^{-3u}(x^2{D}^2+3xD+1){\mid }_{x=u}{e}^{x}du$$$$={\int }_0^{\mathrm{\infty }}{e}^{-3u}(x^2+3x+1)e^x{\mid }_{x=u}du$$$$={\int }_0^{\mathrm{\infty }}{e}^{-2u}(u^2+3u+1)du$$$$=\frac{2}{2^3}+\frac{3}{2^2}+\frac{1}{2}$$$$=\frac{3}{2}$$$$ihavefoundanotherintrestingway.$$

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