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Question Number 43679 by gunawan last updated on 14/Sep/18

$$\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{i^2}{2^i}=...$$

Commented by abdo.msup.com last updated on 14/Sep/18

$$letp\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{x}^{k}with\mid x\mid <1\Rightarrow$$$$p^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}{kx}^{k-1}\Rightarrow {xp}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}k{x}^k$$$$\Rightarrow p^{{}^{\prime }}\left(x\right)+x{p}^{{}^{″}}\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}{k}^2{x}^{k-1}\Rightarrow$$$${xp}^{{}^{\prime }}\left(x\right)+x^2{p}^{{}^{″}}\left(x\right)=\sum _{k=1}^{\mathrm{\infty }}{k}^2{x}^k\Rightarrow$$$$\sum _{k=1}^{\mathrm{\infty }}\frac{k^2}{2^k}=\frac{1}{2}{p}^{{}^{\prime }}\left(\frac{1}{2}\right)+\frac{1}{4}{p}^{″}\left(\frac{1}{2}\right)but$$$$p\left(x\right)=\frac{1}{1-x}\Rightarrow p^{{}^{\prime }}\left(x\right)=\frac{1}{{(1-x)}^2}\Rightarrow$$$$p^{{}^{\prime }}\left(\frac{1}{2}\right)=4also{p}^{{}^{″}}\left(x\right)=-\frac{2(-1)(1-x)}{{(1-x)}^4}$$$$=\frac{2}{{(1-x)}^3}\Rightarrow p^{{}^{″}}\left(\frac{1}{2}\right)=16\Rightarrow$$$$S=\frac{1}{2}\left(4\right)+\frac{1}{4}\left(16\right)=2+4=6.$$

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