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Question Number 79807 by Pratah last updated on 28/Jan/20

Commented by Pratah last updated on 28/Jan/20

$$\mathrm{thanks}\mathrm{sir}$$

Commented by john santu last updated on 28/Jan/20

$$\mathrm{S}=\frac{4}{19}+\frac{44}{{19}^2}+\frac{444}{{19}^3}+\frac{4444}{{19}^4}+...(\ast )$$$$19\mathrm{S}=4+\frac{44}{19}+\frac{444}{{19}^2}+\frac{4444}{{19}^3}+...(\ast \ast )$$$$(\ast \ast )-(\ast )$$$$18\mathrm{S}=4+4\left(\frac{10}{19}\right)+4\left(\frac{100}{{19}^2}\right)+4\left(\frac{1000}{{19}^3}\right)+...$$$$18\mathrm{S}=4\left(\frac{1}{1-\frac{10}{19}}\right)=\frac{4\times 19}{9}$$$$\mathrm{S}=\frac{38}{81}.$$

Answered by mr W last updated on 28/Jan/20

$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{4(1+10+100+...+{10}^{n-1})}{{19}^n}$$$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{4({10}^n-1)}{9\times {19}^n}$$$$S=\frac{4}{9}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{10}^n-1}{{19}^n}$$$$S=\frac{4}{9}\underset{n=1}{\sum ^{\mathrm{\infty }}}[{\left(\frac{10}{19}\right)}^n-{\left(\frac{1}{19}\right)}^n]$$$$S=\frac{4}{9}(\frac{\frac{10}{19}}{1-\frac{10}{19}}-\frac{\frac{1}{19}}{1-\frac{1}{19}})$$$$S=\frac{4}{9}(\frac{10}{9}-\frac{1}{18})$$$$\Rightarrow S=\frac{38}{81}$$

Commented by Pratah last updated on 28/Jan/20

$$\mathrm{thanks}$$

Answered by Smail last updated on 28/Jan/20

$$S=4(\frac{1}{19}+\frac{11}{{19}^2}+\frac{111}{{19}^3}+...)$$$$=4(\frac{1}{19}+\frac{10+1}{{19}^2}+\frac{{10}^2+10+1}{{19}^3}+...)$$$$1+10+{10}^2+...+{10}^n=\frac{1-{10}^{n+1}}{1-10}$$$$S=4(\frac{1-{10}^1}{(1-10)19}+\frac{1-{10}^2}{(1-10){19}^2}+\frac{1-{10}^3}{(1-10){19}^3}+...)$$$$=\frac{4}{1-10}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1-{10}^n}{{19}^n}=\frac{4}{9}(\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{10}{19}\right)}^n-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{19}^n})$$$$=\frac{4}{9}(\frac{10/19}{1-10/19}-\frac{1/19}{1-1/19})$$$$=\frac{4}{9}(\frac{10}{19-10}-\frac{1}{19-1})$$$$S=\frac{38}{81}$$

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