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Question Number 143184 by bramlexs22 last updated on 11/Jun/21

$$\frac{10}{25}+\frac{28}{125}+\frac{82}{625}+...=?$$

Answered by Canebulok last updated on 11/Jun/21

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:$$$$Intermsofsummation,$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\left(3^{n+1}\right)+1}{5^{n+1}}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{3}{5}\right)}^{n+1}+\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{5}\right)}^{n+1}$$$$$$$$Bycalculusmethod,$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^n=\frac{1}{(1-x)}$$$$$$$$Byshiftingtheindexofsummation,$$$$\to Letn=k-1$$$$\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}{x}^{(k-1)}=\frac{1}{(1-x)}$$$$$$$$Bymultiplyingbothsidesby‘‘x^2{}^{″},$$$$\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}{x}^{(k+1)}=\frac{x^2}{(1-x)}$$$$$$$$Letx=\frac{1}{5}$$$$$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{5}\right)}^{n+1}=\frac{{\left(\frac{1}{5}\right)}^2}{(1-\frac{1}{5})}=\frac{1}{20}$$$$$$$$Letx=\frac{3}{5}$$$$$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{3}{5}\right)}^{n+1}=\frac{{\left(\frac{3}{5}\right)}^2}{(1-\frac{3}{5})}=\frac{9}{10}$$$$$$$$Thus;$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\left(3^{n+1}\right)+1}{5^{n+1}}=\frac{1}{20}+\frac{9}{10}=\frac{19}{20}$$$$$$$$\sim Kevin$$

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