Question Number 125663 by Mammadli last updated on 12/Dec/20
$$\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{1}} }+\frac{\mathrm{2}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{6}^{\mathrm{3}} }+…+\frac{{n}}{\mathrm{6}^{{n}} }=? \\ $$
Answered by Dwaipayan Shikari last updated on 12/Dec/20
$${S}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{2}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{6}^{\mathrm{3}} }+..+\frac{{n}}{\mathrm{6}^{{n}} } \\ $$$$−\frac{{S}}{\mathrm{6}}=\:\:\:−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }−\frac{\mathrm{2}}{\mathrm{6}^{\mathrm{3}} }−…−\frac{{n}}{\mathrm{6}^{{n}+\mathrm{1}} } \\ $$$$\frac{\mathrm{5}{S}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+..+\frac{\mathrm{1}}{\mathrm{6}^{{n}} }−\frac{{n}}{\mathrm{6}^{{n}+\mathrm{1}} }\:\Rightarrow{S}=\frac{\mathrm{6}}{\mathrm{5}}\left(\frac{\mathrm{1}−\mathrm{6}^{−{n}} }{\mathrm{5}}−\frac{{n}}{\mathrm{6}^{{n}+\mathrm{1}} }\right)\:\: \\ $$$${S}=\frac{\mathrm{6}−\mathrm{6}^{\mathrm{1}−{n}} }{\mathrm{25}}−\frac{{n}}{\mathrm{5}.\mathrm{6}^{{n}} } \\ $$