Question Number 190521 by Rupesh123 last updated on 04/Apr/23
Answered by mehdee42 last updated on 04/Apr/23
$$\mathrm{8}+\mathrm{88}+\mathrm{888}+...+\mathrm{888}...\mathrm{8}=\mathrm{8}\left(\frac{\mathrm{10}−\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{10}^{\mathrm{2}} −\mathrm{1}}{\mathrm{9}}+...+\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{10}+\mathrm{10}^{\mathrm{2}} +...+\mathrm{10}^{{n}} −{n}\right)=\frac{\mathrm{8}}{\mathrm{9}}\left(\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}}{\mathrm{9}}−{n}\right) \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \frac{\mathrm{81}}{\mathrm{10}^{{n}} }×\frac{\mathrm{8}}{\mathrm{9}}\left(\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}}{\mathrm{9}}−{n}\right)=\mathrm{80} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 04/Apr/23
Excellent, sir!