Sum of Maclaurin series Find the sum of the infinite series
\begin{equation}
\sum_{n=2}^\infty\frac{7n(n-1)}{3^{n-2}}
\end{equation}
I think it probably has something to do with a known Maclaurin series, but cannot for the life of me see which one.. Any hints would be appreciated!
Edit: Using your hints, I was able to solve the problem. Solving it here in case someone is wondering about the same thing:
Manipulate \begin{equation}\sum_{n=2}^\infty x^n=\frac{1}{1-x}-x-1 \end{equation}
by differentiating both sides:
\begin{equation}\sum_{n=2}^\infty nx^{n-1}=\frac{1}{(1-x)^2}-1 \end{equation}
differentiate again:
\begin{equation}\sum_{n=2}^\infty (n-1)nx^{n-2}=\frac{2}{(1-x)^3} \end{equation}
plugging in $\frac{1}{3}$ for x:
\begin{equation}\sum_{n=2}^\infty n(n-1)(\frac{1}{3})^{n-2}=\frac{2}{(1-\frac{1}{3})^3} \end{equation}
which is equivalent to
\begin{equation}\sum_{n=2}^\infty \frac{n(n-1)}{3^{n-2}} \end{equation}
finally, multiplying by 7 gives us
\begin{equation}\begin{split}\sum_{n=2}^\infty \frac{7n(n-1)}{3^{n-2}}&=7\frac{2}{(1-\frac{1}{3})^3}\\
&=\underline{\underline{\frac{189}{4}}} \end{split} \end{equation}
 A: Hint: Notice 
$$ \sum x^n = \frac{ 1}{1- x} $$
$$ \sum n x^{n-1} = \frac{1}{(1-x)^2} $$
$$ \sum n (n-1) x^{n-2} = \frac{2 }{(1-x)^3} $$
they all converge for $|x| < 1$
A: Hint. We have that for $x\not=1$, and $N\geq 2$,
$$\frac{d^2}{dx^2}\left(\frac{1-x^{N+1}}{1-x}\right)=\frac{d^2}{dx^2}\left(\sum_{n=0}^N x^n\right)=\sum_{n=2}^N n(n-1)x^{n-2}.$$
P.S. for the downvoters. I considered a finite sum because it is not so straightforward to say that we can interchange the differentiation and the infinite sum: for $|x|<1$,
$$\frac{d^2}{dx^2}\left(\frac{1}{1-x}\right)=\frac{d^2}{dx^2}\left(\sum_{n=0}^{\infty} x^n\right)=
\sum_{n=0}^{\infty} \frac{d^2(x^n)}{dx^2}
=\sum_{n=2}^{\infty} n(n-1)x^{n-2}.$$
A: You want
$\sum_{n=2}^{\infty} n(n-1) x^n
$
for a certain value of $x$.
Start with
$\sum_{n=2}^{\infty} x^n
=\dfrac{x^2}{1-x}
$
and differentiate it twice.
A: You can't just differentiate an infinite series term by term and expect the result to be the derivative. It is true that for a power series this process yields the expected result with the same radius of convergence, but it is a non-trivial fact.
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
Furthermore, there is no need for differentiation or integration here.
Let $S(m) = \sum_{n=2}^m \lfrac{7n(n-1)}{3^{n-2}}$.
Then $S(m) - \lfrac13 S(m) = \sum_{n=2}^m \lfrac{7n(n-1)}{3^{n-2}} - \sum_{n=3}^{m+1} \lfrac{7(n-1)(n-2)}{3^{n-2}}$
  $\ = \sum_{n=2}^m \lfrac{7n(n-1)}{3^{n-2}} - \sum_{n=2}^{m+1} \lfrac{7(n-1)(n-2)}{3^{n-2}}$   [This step is not needed but convenient.]
  $\ = \sum_{n=2}^{m+1} \lfrac{14(n-1)}{3^{n-2}} - \lfrac{7(m+1)m}{3^{m-1}}$.
This reduces the numerator polynomial degree by one. Repeat this one more time and you will get a usual geometric series, whose sum you could also prove by the same method. At the end just take $m \to \infty$ to get the answer.
