Generating function: Find a closed form of $\sum_{k=0}^n (-3)^k(k+1)$

Question

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!

Answer 1

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 $-5+3*(-3)^2 = -5+27 = 22$.

Answer 2

We have that $$A(x)=\sum_{n=0}^\infty a_nx^n,\qquad a_n=\sum_{k=0}^n(-3)^k(k+1).$$ Note that $$a_{n+1}-a_n=(-3)^{n+1}(n+2),$$ which implies that $$\begin{align*} (1-x)A(x)=A(x)-xA(x)&=\sum_{n=0}^\infty (a_nx^{n}-a_nx^{n+1})\\\\ &=a_0+\sum_{n=1}^\infty (a_n-a_{n-1})x^n\\\\\\ &=1+\sum_{n=1}^\infty(-3)^{n+1}(n+2)x^n\\\\\\ &=1-3\sum_{n=1}^\infty n(-3x)^n-6\sum_{n=1}^\infty (-3x)^n \end{align*}$$ Can you take it the rest of the way?

Answer 3

Let $a_{n}=(-3)^n(n+1)$ and let $f(x)=\sum_n a_{n}x^n$ be the generating function of $a_{n}$.

From a property of generating functions we have,that if $f$ is the Generating Function of {$a_{n}$},then $\frac{f}{1-x}$ is the generating function of {$\sum_{k=0}^n a_{k}$}$_{n\ge0}$,which is precisely what you want.

Now,in order to calculate $f$,we have $$\sum_{n\ge0}a_{n}x^n=\sum_{n\ge0} n(-3x)^n+\sum_{n\ge0} (-3x)^n$$

$$\Rightarrow f=\frac{-3x}{(1+3x)^2} + \frac{1}{1+3x}=\frac{1}{(1+3x)^2}$$

Therefore,$$\frac{f}{1-x}=\frac{1}{(1-x)(1+3x)^2}$$Which is the required generating function.