Question

Álgebra...............​

Answer 1

$\displaystyle\\\sum_{n=1}^{k}\dfrac{n}{3^n}-\dfrac{n-1}{3^{n-1}}=\dfrac{k}{3^k}-\dfrac{-1}{3^{-1}}\\ \\ \\\sum_{n=1}^{k}\dfrac{1}{3^n}\left(n-\dfrac{n-1}{3^{-1}}\right) =\dfrac{k}{3^k}+3\\ \\ \\\sum_{n=1}^{k}\dfrac{1}{3^n}(-2n+3)=\dfrac{k}{3^k}+3\\ \\ \\-2\sum_{n=1}^{k}\dfrac{n}{3^n}+3\sum_{n=1}^{k}\dfrac{1}{3^n}=\dfrac{k}{3^k}+3\\ \\ \\-2\sum_{n=1}^{\infty}\dfrac{n}{3^n}+3\sum_{n=1}^{\infty}\dfrac{1}{3^n}=3$  

$\displaystyle\\-2\sum_{n=1}^{\infty}\dfrac{n}{3^n}+3\cdot \dfrac{3}{2}=3\\ \\ \\-2\sum_{n=1}^{\infty}\dfrac{n}{3^n}+ \dfrac{9}{2}=3\\ \\ \\-2\sum_{n=1}^{\infty}\dfrac{n}{3^n}= \dfrac{-3}{2}\\ \\ \\\sum_{n=1}^{\infty}\dfrac{n}{3^n}= \dfrac{3}{4}$