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Question Number 33088 by Joel578 last updated on 10/Apr/18

$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2}{2^{n-1}}$$

Commented by abdo imad last updated on 10/Apr/18

$$letconsidertheseries\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{x}^{n}with\mid x\mid <1$$$$wehaves\left(x\right)=\frac{1}{1-x}\Rightarrow s^{{}^{\prime }}\left(x\right)=\frac{1}{{(1-x)}^2}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}n{x}^{n-1}$$$$=\frac{1}{{(1-x)}^2}also{s}^{{}^{″}}\left(x\right)=-\frac{2(-1)(1-x)}{{(1-x)}^4}=\frac{2}{{(1-x)}^3}\Rightarrow$$$$\sum _{n=2}^{\mathrm{\infty }}n(n-1)x^{n-2}=\frac{2}{{(1-x)}^3}\Rightarrow$$$$\sum _{n=2}^{\mathrm{\infty }}n(n-1)x^{n-1}=\frac{2x}{{(1-x)}^3}\Rightarrow$$$$\sum _{n=2}^{\mathrm{\infty }}{n}^2{x}^{n-1}=\frac{2x}{{(1-x)}^3}+\sum _{n=2}^{\mathrm{\infty }}{nx}^{n-1}$$$$=\frac{2x}{{(1-x)}^3}+\frac{1}{{(1-x)}^2}-1\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{n}^2{x}^{n-1}=\frac{2x}{{(1-x)}^3}+\frac{1}{{(1-x)}^2}$$$$afterwetakex=\frac{1}{2}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{2^{n-1}}=\frac{1}{{\left(\frac{1}{2}\right)}^3}+\frac{1}{{\left(\frac{1}{2}\right)}^2}=8+4=12.$$$$$$

Commented by abdo imad last updated on 10/Apr/18

$$remark★wehave2\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{2^n}=12\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{2^n}=6.$$

Commented by Joel578 last updated on 13/Apr/18

$$Thankyouverymuch$$

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