Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 159379 by HongKing last updated on 16/Nov/21

$$\mathrm{let}\mathbf{S}\left(\mathrm{x}\right)=\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}{\left(3\mathrm{x}\right)}^{\mathbf{n}+2}$$$$\mathrm{using}\mathrm{the}\mathrm{sum}\mathrm{above}\mathrm{find}:$$$$\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathbf{n}+1}}{3^{\mathbf{n}+1}(\mathrm{n}+3)}$$

Answered by Ar Brandon last updated on 16/Nov/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-\frac{1}{3})}^{n+1}\frac{1}{n+3}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-\frac{1}{3})}^{n+1}{\int }_0^1{x}^{n+2}dx$$$$=-3\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-\frac{1}{3})}^{n+2}{\int }_0^1{x}^{n+2}dx=-3{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-\frac{x}{3})}^{n+2}dx$$$$=-3{\int }_0^1\frac{\frac{x^2}{9}}{1+\frac{x}{3}}dx=-{\int }_0^1\frac{x^2}{3+x}dx=-{\int }_0^1\frac{{(x+3)}^2-6(x+3)+9}{x+3}dx$$$$=-{[\frac{x^2}{2}+3x-6x+9\ln (x+3)]}_0^1=9\ln 3+\frac{5}{2}-9\ln 4=\frac{5}{2}+9\ln \left(\frac{3}{4}\right)$$

Commented by HongKing last updated on 16/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathbf{S}\mathrm{er}\mathrm{cool}$$

Commented by HongKing last updated on 16/Nov/21

$$\mathrm{my}\mathrm{dear}\mathrm{Ser},\mathrm{but}\mathrm{they}\mathrm{said}\mathrm{use}\mathrm{the}\mathrm{first}$$$$\mathrm{sum}\mathrm{S}\left(\mathrm{x}\right)\mathrm{to}\mathrm{evaluate}\mathrm{the}\mathrm{new}\mathrm{one}$$

Commented by Ar Brandon last updated on 16/Nov/21

Answered by Ar Brandon last updated on 16/Nov/21

$$S\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(3x\right)}^{n+2}=\frac{9x^2}{1-3x}=\frac{{(1-3x)}^2-2(1-3x)+1}{1-3x}$$$$\int S\left(x\right)dx=\frac{1}{3}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(3x\right)}^{n+3}}{n+3}+C=x-\frac{3}{2}{x}^2-2x-\frac{\ln (1-3x)}{3}+C$$$$\int S\left(0\right)dx=0=C\Rightarrow \int S\left(x\right)dx=\frac{1}{3}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(3x\right)}^{n+3}}{n+3}=-x-\frac{3}{2}{x}^2-\frac{\ln (1-3x)}{3}$$$$\int S(-\frac{1}{9})dx=\frac{1}{3}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+3}}{3^{n+3}(n+3)}=\frac{1}{9}-\frac{1}{54}-\frac{1}{3}\ln \left(\frac{4}{3}\right)$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{3^{n+1}(n+3)}=3-\frac{1}{2}-9\ln \left(\frac{4}{3}\right)=\frac{5}{2}-9\ln \left(\frac{4}{3}\right)$$

Commented by HongKing last updated on 16/Nov/21

$$\mathrm{perfect},\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Ser}$$

Commented by Ar Brandon last updated on 16/Nov/21

You're welcome, Sir.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com