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Question Number 30049 by abdo imad last updated on 15/Feb/18

$$find\sum _{n=1}^{\mathrm{\infty }}\frac{n(n+1)}{3^n}.$$

Commented by prof Abdo imad last updated on 16/Feb/18

$$wehaveprovedthat\sum _{n=0}^{\mathrm{\infty }}(n+1)x^n=\frac{1}{{(1-x)}^2}$$$$for\mid x\mid <1duetouniformconvergenceon[0,1]$$$$byderivation\sum _{n=1}^{\mathrm{\infty }}n(n+1)x^{n-1}=\frac{d}{dx}\left({(1-x)}^{-2}\right)$$$$=(-2)(-1){(1-x)}^{-3}=\frac{2}{{(1-x)}^3}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}n(n+1)x^n=\frac{2x}{{(1-x)}^3}andforx=\frac{1}{3}weget$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{n(n+1)}{3^n}=\frac{\frac{2}{3}}{{\left(\frac{2}{3}\right)}^3}=\frac{1}{{\left(\frac{2}{3}\right)}^2}=\frac{9}{4}=.$$$$$$

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