Question

porfa hagan la 8 la necesito

Answer 1

Veamos la sucesión

6, 18, 72, 324

6, 6^2/2 , 6^3 / 3, 6^4 / 4... 6^n / n

$\displaystyle\\M=\sum_{n=1}^{\infty} \dfrac{n}{6^n}\\ \\ \\\sum_{n=1}^{j}\dfrac{n}{6^{n}}-\dfrac{n+1}{6^{n+1}} = \dfrac{1}{6}-\dfrac{j+1}{6^{j+1}}\\ \\ \\\sum_{n=1}^{j}\dfrac{1}{6^n}(\dfrac{5n-1}{6}) = \dfrac{1}{6}-\dfrac{j+1}{6^{j+1}}\\ \\ \\\dfrac{5}{6}\sum_{n=1}^{j}\dfrac{n}{6^n}-\dfrac{1}{6}\sum_{n=1}^{j}\dfrac{1}{6^n}= \dfrac{1}{6}-\dfrac{j+1}{6^{j+1}}\\ \\ \\5\sum_{n=1}^{j}\dfrac{n}{6^n}-\sum_{n=1}^{j}\dfrac{1}{6^n}=1-\dfrac{j+1}{6^{j}}$

$\displaystyle\\5\sum_{n=1}^{\infty}\dfrac{n}{6^n}-\sum_{n=1}^{\infty}\dfrac{1}{6^n}=1\\ \\ \\5\sum_{n=1}^{\infty}\dfrac{n}{6^n} - \dfrac{1}{5}=1\\ \\ \\5\sum_{n=1}^{\infty}\dfrac{n}{6^n} =\dfrac{6}{5} \\ \\ \\\sum_{n=1}^{\infty}\dfrac{n}{6^n} =\dfrac{6}{25}$