Question

Quien sabe cómo se resuelve este problema:(​

Answer 1

Respuesta:

$\begin{lgathered}\displaystyle M = \frac{216}{25} \end{lgathered}$

Explicación paso a paso

$\begin{lgathered}\displaystyle Si: \: |x| > 1 \end{lgathered}$

$\begin{lgathered}\displaystyle \\\sum_{n=1}^{\infty} \frac{1}{x^n} = \frac{1}{x-1} \end{lgathered}$

$\begin{lgathered}\displaystyle \\\sum_{n=1}^{\infty} \frac{n}{x^n} = \frac{x}{(x-1)^2} \end{lgathered}$

Resuelve:

$\begin{lgathered}\displaystyle M = 7 + \frac{8}{6} + \frac{9}{ {6}^{2} } + \frac{10}{ {6}^{3} } + \cdots \end{lgathered}$

$\begin{lgathered}\displaystyle \\\sum_{n=1}^{\infty} \frac{n+6}{6^{n-1}} = \sum_{n=1}^{\infty} \frac{n}{6^{n-1}} + \frac{6}{ {6}^{n - 1} } \end{lgathered}$

$\begin{lgathered}\displaystyle \\\sum_{n=1}^{\infty} \frac{n}{6^{n-1}} + 6\sum_{n=1}^{\infty} \frac{1}{ {6}^{n - 1} } \end{lgathered}$

$\begin{lgathered}\displaystyle \\6\sum_{n=1}^{\infty} \frac{n}{6^{n}} + 36\sum_{n=1}^{\infty} \frac{1}{ {6}^{n}} \end{lgathered}$

$\begin{lgathered}\displaystyle 6 \left ( \frac{6}{(6-1)^2 } \right )+ 36 \left ( \frac{1}{6-1} \right )\end{lgathered}$

$\begin{lgathered}\displaystyle \left ( \frac{36}{25 } \right )+ \left ( \frac{36}{5} \right )\end{lgathered}$

$\begin{lgathered}\displaystyle \frac{36}{25 } + \frac{180}{25} \end{lgathered}$

$\begin{lgathered}\displaystyle \frac{216}{25} \end{lgathered}$