help solving a geometric sequence after manipulation Evaluate the sum. $\sum_{n=2}^\infty \frac{n(-3)^n}{4^{n+1}}$
I have manipulated it to be in a more common form I thought I could solve but I'm still having trouble getting the correct sum. I'm using wolfram to check against. It says the sum evaluates to $\frac{99}{784}=0.12628$
Here is what I have so far.
$\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n+1}}$
$=\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n}\cdot4}$
$=\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n}}\ \cdot\frac{1}{4}$
$=\sum_{n=2}^{\infty}n\left(-\frac{3}{4}\right)^{n}\ \cdot\frac{1}{4}$
$=\frac{1}{4}\sum_{n=2}^{\infty}n\left(-\frac{3}{4}\right)^{n}$
was trying to get the common difference to solve that way but it's not coming out right. How do I properly evaluate the remaining sum?

$1+x+x^2+x^3+...=\frac{1}{1-x} \iff \sum_{n\geq0}x^n=\frac{1}{1-x}\Rightarrow \sum_{n\geq1}nx^{n-1}=\frac{1}{(1-x)^2}$
$\sum_{n\geq2}n\Big(-\frac{3}{4}\Big)^{n}=-\frac{3}{4}\sum_{n\geq2}n\Big(-\frac{3}{4}\Big)^{n-1}=-\frac{3}{4} \Big( \frac{1}{(1+\frac{3}{4})^2}-1\Big)$
