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Question Number 46852 by maxmathsup by imad last updated on 01/Nov/18

$$findthevalueof\sum _{n=1}^{\mathrm{\infty }}\frac{n^3}{3^n}$$

Commented by maxmathsup by imad last updated on 02/Nov/18

$$wehaveprovedthatfor\mid x\mid <1\sum _{n=1}^{\mathrm{\infty }}{nx}^n=\frac{x}{{(1-x)}^2}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}{n}^2{x}^{n-1}=\frac{{(1-x)}^2+2x(1-x)}{{(1-x)}^4}=\frac{1-x+2x}{{(1-x)}^3}=\frac{x+1}{{(1-x)}^3}\Rightarrow$$$$n^2{x}^n=\frac{x^2+x}{{(1-x)}^3}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{n}^3{x}^{n-1}=\frac{(2x+1){(1-x)}^3+3{(1-x)}^2(x^2+x)}{{(1-x)}^6}$$$$=\frac{(2x+1)(1-x)+3(x^2+x)}{{(1-x)}^4}=\frac{2x-2x^2+1-x+3x^2+3x}{{(1-x)}^4}=\frac{x^2+4x+1}{{(1-x)}^4}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}{n}^3{x}^n=\frac{x^3+4x^2+x}{{(1-x)}^4}$$$$x=\frac{1}{3}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{n}^3{\left(\frac{1}{3}\right)}^n=\frac{\frac{1}{27}+\frac{4}{9}+\frac{1}{3}}{{\left(\frac{2}{3}\right)}^4}=\frac{1}{27}\frac{1+12+9}{\frac{2^4}{3^4}}=\frac{1}{27}\frac{3^4}{2^4}.22$$$$=\frac{3}{8.2}.\frac{11.2}{1}=\frac{33}{8}\Rightarrow ★\sum _{n=1}^{\mathrm{\infty }}\frac{n^3}{3^n}=\frac{33}{8}★$$

Answered by Smail last updated on 02/Nov/18

$$\frac{1}{1-x}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^{n}with\mid x\mid <1$$$$\frac{1}{{(1-x)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^{n-1}$$$$\frac{x}{{(1-x)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^n$$$$-\frac{1+x}{{(1-x)}^3}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^2{x}^{n-1}\iff \frac{-x(1+x)}{{(1-x)}^3}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^2{x}^n$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^3{x}^{n-1}=\frac{(2x+1)(1-x)+3x(1+x)}{{(1-x)}^4}$$$$=\frac{x^2+4x+1}{{(1-x)}^4}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^3{x}^n=\frac{x(x^2+4x+1)}{{(1-x)}^4}$$$$x=\frac{1}{3}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^3}{3^n}=\frac{\frac{1}{9}+\frac{4}{3}+1}{3{\left(\frac{2}{3}\right)}^4}=\frac{33}{8}$$

Commented by maxmathsup by imad last updated on 02/Nov/18

$$correctanswerthankssir.$$

Commented by Smail last updated on 02/Nov/18

$$Thankyouforpostingquistionslikethis.$$

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