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Question Number 125663 by Mammadli last updated on 12/Dec/20

$$\frac{1}{6^1}+\frac{2}{6^2}+\frac{3}{6^3}+...+\frac{n}{6^n}=?$$

Answered by Dwaipayan Shikari last updated on 12/Dec/20

$$S=\frac{1}{6}+\frac{2}{6^2}+\frac{3}{6^3}+..+\frac{n}{6^n}$$$$-\frac{S}{6}=-\frac{1}{6^2}-\frac{2}{6^3}-...-\frac{n}{6^{n+1}}$$$$\frac{5S}{6}=\frac{1}{6}+\frac{1}{6^2}+..+\frac{1}{6^n}-\frac{n}{6^{n+1}}\Rightarrow S=\frac{6}{5}(\frac{1-6^{-n}}{5}-\frac{n}{6^{n+1}})$$$$S=\frac{6-6^{1-n}}{25}-\frac{n}{5.6^n}$$

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