Question Number 129409 by Algoritm last updated on 15/Jan/21
Commented by soumyasaha last updated on 15/Jan/21
![= (4/9)[ 1+(1/(9.11))+(1/(9^2 .11^2 )) + (1/(9^3 .11^3 ))+...] = (4/9)[ 1+(1/(99))+((1/(99)))^2 +((1/(99)))^3 +...] = (4/9).(1/(1−(1/(99)))) = (4/9).((99)/(98)) = ((22)/(49))](https://www.tinkutara.com/question/Q129412.png)
$$\:\:\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{9}.\mathrm{11}}+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} .\mathrm{11}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{3}} .\mathrm{11}^{\mathrm{3}} }+…\right] \\ $$$$\:\:\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{99}}+\left(\frac{\mathrm{1}}{\mathrm{99}}\right)^{\mathrm{2}} \:+\left(\frac{\mathrm{1}}{\mathrm{99}}\right)^{\mathrm{3}} +…\right] \\ $$$$\:\:\:=\:\frac{\mathrm{4}}{\mathrm{9}}.\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{99}}} \\ $$$$\:\:\:=\:\frac{\mathrm{4}}{\mathrm{9}}.\frac{\mathrm{99}}{\mathrm{98}}\: \\ $$$$\:\:\:=\:\frac{\mathrm{22}}{\mathrm{49}}\: \\ $$
Answered by Olaf last updated on 15/Jan/21
![S = (4/9)+((44)/(99^2 ))+((444)/(999^3 ))+... S = (4/9)+((4×11)/(9^2 ×11^2 ))+((4×111)/(9^3 ×111^3 ))+... S = (4/9)[1+(1/(9×11))+(1/(9^2 ×111^2 ))+...] {1, 11, 111, 1111...} = ((10^n −1)/9), n ≥ 1 S = (4/9)Σ_(n=0) ^∞ (1/(9^n (((10^(n+1) −1)/9))^n )) S = (4/9)Σ_(n=0) ^∞ (1/((10^(n+1) −1)^n )) ...to be continued...](https://www.tinkutara.com/question/Q129427.png)
$$ \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{44}}{\mathrm{99}^{\mathrm{2}} }+\frac{\mathrm{444}}{\mathrm{999}^{\mathrm{3}} }+… \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{4}×\mathrm{11}}{\mathrm{9}^{\mathrm{2}} ×\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{4}×\mathrm{111}}{\mathrm{9}^{\mathrm{3}} ×\mathrm{111}^{\mathrm{3}} }+… \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{9}×\mathrm{11}}+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} ×\mathrm{111}^{\mathrm{2}} }+…\right] \\ $$$$\left\{\mathrm{1},\:\mathrm{11},\:\mathrm{111},\:\mathrm{1111}…\right\}\:=\:\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{9}},\:{n}\:\geqslant\:\mathrm{1} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{4}}{\mathrm{9}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{9}^{{n}} \left(\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{9}}\right)^{{n}} } \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{4}}{\mathrm{9}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1}\right)^{{n}} } \\ $$$$…{to}\:{be}\:{continued}… \\ $$