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Question Number 106488 by Algoritm last updated on 05/Aug/20

Commented by Dwaipayan Shikari last updated on 05/Aug/20

$$Q106379$$$$\frac{1.2^2}{10}+\frac{2.3^2}{10}+.....=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n{(n+1)}^2}{{10}^n}$$

Answered by abdomsup last updated on 05/Aug/20

$$S=\sum _{n=1}^{\mathrm{\infty }}\frac{n{(n+1)}^2}{{10}^n}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{n(n^2+2n+1)}{{10}^n}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{n^3}{{10}^n}+2\sum _{n=0}^{\mathrm{\infty }}\frac{n^2}{{10}^n}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{10}^n}$$$$wehavefor\mid x\mid <1$$$$\sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{{10}^n}$$$$=\frac{1}{1-\frac{1}{10}}=\frac{10}{9}$$$$also\sum _{n=1}^{\mathrm{\infty }}{nx}^{n-1}=\frac{1}{{(1-x)}^2}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}{nx}^n=\frac{x}{{(1-x)}^2}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}{n}^2{x}^{n-1}=\frac{{(x-1)}^2-2(x-1)x}{{(x-1)}^4}$$$$=\frac{x-1-2x}{{(x-1)}^3}=\frac{-x-1}{{(x-1)}^3}=\frac{1+x}{{(1-x)}^3}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}{n}^2{x}^n=\frac{x+x^2}{{(1-x)}^3}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{{10}^n}=\frac{\frac{1}{10}+{\left(\frac{1}{10}\right)}^2}{{(1-\frac{1}{10})}^3}=...$$$$\sum _{n=1}^{\mathrm{\infty }}{n}^3{x}^{n-1}=\frac{(2x+1){(1-x)}^3+3{(1-x)}^2(x^2+x)}{{(1-x)}^6}$$$$=\frac{(2x+1)(1-x)+3x^2+3x}{{(1-x)}^4}$$$$=\frac{2x+1-2x^2-x+3x^2+3x}{{(1-x)}^4}$$$$=\frac{x^2+4x+1}{{(1-x)}^4}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}{n}^3{x}^n=\frac{x^3+4x^2+x}{{(1-x)}^4}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^3}{{10}^n}=\frac{\frac{1}{{10}^3}+\frac{4}{{10}^2}+\frac{1}{10}}{{(1-\frac{1}{10})}^4}=...$$

Answered by Dwaipayan Shikari last updated on 05/Aug/20

$$\sum ^{\mathrm{\infty }}\frac{n{(n+1)}^2}{{10}^n}$$$$\sum ^{\mathrm{\infty }}\frac{n^3}{{10}^n}+\sum ^{\mathrm{\infty }}\frac{2n^2}{{10}^n}+\sum ^{\mathrm{\infty }}\frac{n}{{10}^n}$$$$\sum ^{\mathrm{\infty }}\frac{n}{{10}^n}=\frac{1}{10}+\frac{2}{{10}^2}+...=S$$$$\frac{S}{10}=\frac{1}{{10}^2}+\frac{2}{{10}^3}+....$$$$\frac{9S}{10}=\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+...$$$$\frac{9S}{10}=\frac{\frac{1}{10}}{1-\frac{1}{10}}=\frac{1}{9}$$$$S=\frac{10}{81}$$$$2\sum ^{\mathrm{\infty }}\frac{n^2}{{10}^n}=\frac{1}{10}+\frac{4}{{10}^2}+\frac{9}{{10}^2}+....=S^{\prime }$$$$\frac{S^{\prime }}{10}=\frac{1}{{10}^2}+\frac{4}{{10}^3}+\frac{9}{{10}^4}+...$$$$\frac{9S^{\prime }}{10}=\frac{1}{10}+\frac{3}{{10}^2}+\frac{5}{{10}^3}+...$$$$\frac{9S^{{}^{\prime }}}{100}=\frac{1}{{10}^2}+\frac{3}{{10}^3}+...$$$$......subtracting$$$$\frac{9S^{\prime }}{10}(1-\frac{1}{10})=\frac{1}{10}+2(\frac{1}{{10}^2}+.....)$$$$\frac{81S^{\prime }}{100}=\frac{2}{9}-\frac{1}{10}$$$$S^{\prime }=\frac{10}{9}.\frac{11}{81}=\frac{110}{729}$$$$\sum ^{\mathrm{\infty }}\left(\frac{n^3}{{10}^n}\right)=\frac{1}{10}+\frac{8}{{10}^2}+\frac{27}{{10}^3}+...=S_n$$$$S_n+2S^{\prime }+S=\sum ^{\mathrm{\infty }}\frac{n^3}{{10}^n}+\frac{220}{729}+\frac{10}{81}=\frac{1400}{2187}$$

Answered by JDamian last updated on 06/Aug/20

$$S=\underset{k=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{k{(k+1)}^2}{{10}^k}$$$$f\equiv f\left(x\right)=1+x^2+x^3+\cdot \cdot \cdot =\frac{1}{1-x}\mathrm{\forall }\mid x\mid <1$$$$D\equiv \frac{d}{dx}$$$$Df=1+2x+3x^2+\cdot \cdot \cdot =\frac{1}{{(1-x)}^2}\mathrm{\forall }\mid x\mid <1$$$$x^{2}Df=1\cdot x^2+2x^3+\cdot \cdot \cdot =\frac{x^2}{{(1-x)}^2}\mathrm{\forall }\mid x\mid <1$$$$D\left(x^{2}Df\right)=1\cdot 2x+2\cdot 3x^2+3\cdot 4x^3+\cdot \cdot \cdot =$$$$=2\frac{x}{{(1-x)}^3}\mathrm{\forall }\mid x\mid <1$$$$xD\left(x^{2}Df\right)=1\cdot 2x^2+2\cdot 3x^3+3\cdot 4x^4+\cdot \cdot \cdot =$$$$=2\frac{x^2}{{(1-x)}^3}\mathrm{\forall }\mid x\mid <1$$$$g\equiv D\left(xD\left(x^{2}Df\right)\right)=1\cdot 2^{2}x+2\cdot 3^2{x}^2+$$$$+3\cdot 4^2{x}^3+4\cdot 5^2{x}^4+\cdot \cdot \cdot =$$$$=2\frac{(2+x)x}{{(1-x)}^4}\mathrm{\forall }\mid x\mid <1$$$$$$$$g\left(\frac{1}{10}\right)=2\frac{(2+\frac{1}{10})\frac{1}{10}}{{\left(\frac{9}{10}\right)}^4}=2\frac{\frac{21}{100}}{{\left(\frac{9}{10}\right)}^4}=\frac{4200}{9^4}=\frac{4200}{6561}$$

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