Generating function: Find a closed form of $\sum_{k=0}^n (-3)^k(k+1)$ Find the closed form of $\sum_{k=0}^n (-3)^k(k+1)$.
So the generating function would be: $$A(x)=1-6x+18x^2-108x^3...$$ So what I did notice is that its closed form is perhaps some variation of $1\over {1+x}$ but I didn't manage to find a general formula. 
Thanks for any help in advance! 

Let $f(x) = \sum_{k=0}^n x^{k+1}
= \frac{x-x^{n+2}}{1-x}
$.
Then $f'(x) = \sum_{k=0}^n (k+1)x^k$,
so $f'(-3) = \sum_{k=0}^n (k+1)(-3)^k$.
So 
$\begin{align}
f'(x) &= \frac{(x-x^{n+2})'(1-x) - (1-x)'(x-x^{n+2})}{(1-x)^2}\\
&= \frac{(1-x)(1-(n+2)x^{n+1})+x-x^{n+2}}{(1-x)^2}\\
&= \frac{1-(n+2)x^{n+1}-x+(n+2)x^{n+2}+x-x^{n+2}}{(1-x)^2}\\
&= \frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(1-x)^2}\\
\end{align}
$
Putting $x = -3$,
$\begin{align}
f'(-3) 
&= \frac{1-(n+2)(-3)^{n+1}+(n+1)(-3)^{n+2}}{4^2}\\
&= \frac{1-(-3)^{n+1}((n+2)+(n+1)3)}{4^2}\\
&= \frac{1-(-3)^{n+1}(4n+5)}{16}\\
\end{align}
$
As a check (since I did this off the top of my head):
If $n=0$, 
$f'(-3) = \frac{1-(-3)^{1}(5)}{16}
=\frac{1+15}{16}
=1
$
and $1*(-3)^0 = 1$.
If $n=1$, 
$f'(-3) = \frac{1-(-3)^{2}(4+5)}{16}
=\frac{1-9*9}{16}
=-5
$
and $1+2*(-3) = -5$.
If $n=2$, 
$f'(-3) = \frac{1-(-3)^{3}(8+5)}{16}
=\frac{1+27*13}{16}
=22
$
and $-5+3*(-3)^2 = -5+27 = 22$.
