Reduce this series Let $$f(x)= \sum_{n=2}^\infty nx^n4^n$$
How do we reduce this?
I know that $$\sum_{n=0}^\infty nx^n = \frac{x}{(1-x)^2}$$ and $$\sum_{n=0}^\infty a^n x^n = \frac{1}{1-ax}$$
But how do I combine both?
 A: You already know
$$
\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}\tag{1}
$$
(the $n=0$ term is $0$)
Just substitute $x\mapsto4x$ in $(1)$ to get
$$
\sum_{n=1}^\infty n(4x)^n=\frac{4x}{(1-4x)^2}\tag{2}
$$
then subtract the $n=1$ term from both sides
$$
\sum_{n=2}^\infty n4^nx^n=\frac{4x}{(1-4x)^2}-4x\tag{3}
$$
You can simplify $(3)$ however seems best
Note that since $(1)$ converges for $|x|\lt1$, $(3)$ converges for $|4x|\lt1$, or $|x|\lt\frac14$.
A: i have the partial sum and it looks like the following
${\frac { \left( 4\,x \right) ^{m+1} \left( 4\, \left( m+1 \right) x-m-
1-4\,x \right) }{ \left( 4\,x-1 \right) ^{2}}}-16\,{\frac {{x}^{2}
 \left( 4\,x-2 \right) }{ \left( 4\,x-1 \right) ^{2}}}
$
A: 
Here's a way to find a closed expression for this sum
\begin{align*}
f(x)&=\sum_{n=2}^{\infty}nx^n4^n\\
&=\sum_{n=2}^{\infty}n(4x)^n\\
&=4x\sum_{n=2}^{\infty}n(4x)^{n-1}\\
&=x\frac{d}{dx}\sum_{n=2}^{\infty}(4x)^{n}\tag{1}\\
&=x\left(\frac{d}{dx}\left(\frac{1}{1-4x}-1-4x\right)\right)\tag{2}\\
&=x\left(\frac{4}{(1-4x)^2}-4\right)\tag{3}\\
&=4x\left(\frac{1}{(1-4x)^2}-1\right)
\end{align*}

Observe that 


*

*in (1) we use $\frac{d}{dx}(4x)^n=4n(4x)^{n-1}$

*in (2) we use the geometric series $\sum_{n=0}^{\infty}(4x)^n=\frac{1}{1-4x}$

*in (3) we differentiate $\frac{1}{1-4x}$ to get $\frac{4}{(1-4x)^2}$
