Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 106379 by Algoritm last updated on 04/Aug/20

Answered by mr W last updated on 04/Aug/20

$$\frac{x^2}{1-x}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{n+1}for\mid x\mid <1$$$$\frac{2x}{1-x}+\frac{x^2}{{(1-x)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(n+1)x^n$$$$\frac{2x^2}{1-x}+\frac{x^3}{{(1-x)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(n+1)x^{n+1}$$$$\frac{4x}{1-x}+\frac{5x^2}{{(1-x)}^2}+\frac{2x^3}{{(1-x)}^3}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(n+1)}^2{x}^n$$$$\frac{4}{1-x}+\frac{14x}{{(1-x)}^2}+\frac{16x^2}{{(1-x)}^3}+\frac{6x^3}{{(1-x)}^4}=\underset{n=1}{\sum ^{\mathrm{\infty }}}n{(n+1)}^2{x}^{n-1}$$$$\frac{4x}{1-x}+\frac{14x^2}{{(1-x)}^2}+\frac{16x^3}{{(1-x)}^3}+\frac{6x^4}{{(1-x)}^4}=\underset{n=1}{\sum ^{\mathrm{\infty }}}n{(n+1)}^2{x}^n$$$$setx=\frac{1}{10}<1$$$$\frac{4}{9}+\frac{14}{9^2}+\frac{16}{9^3}+\frac{6}{9^4}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n{(n+1)}^2}{{10}^n}$$$$\frac{6+16\times 9+14\times 9^2+4\times 9^3}{9^4}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n{(n+1)}^2}{{10}^n}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n{(n+1)}^2}{{10}^n}=\frac{1400}{2187}$$

Commented by Algoritm last updated on 05/Aug/20

$$\mathrm{thank}\mathrm{you}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com