Consider the following series $$\sum_{k=0}^{\infty}\frac{(k+1)(k+3)(-1)^k}{3^k}$$
I'm told to analytically find the sum to infinity and I have been given this as a clue.
$$\Sigma_{k=0}^\infty x^k = \frac{1}{1-x} \text{ if } |x|<1$$
I know that the answer is $\frac{45}{32}$ but that's just because of wolfram. I really have no idea where to begin.
I tried to write it out and find a pattern but I could not spot one and I don't understand how the hint is useful here.
Best Answer
$$\frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^kx^k$$ multiply by $x$ $$\frac{x}{1+x}=\sum_{k=0}^{\infty}(-1)^kx^{k+1}$$
$$(\frac{x}{1+x})'=\sum_{k=0}^{\infty}(-1)^k(k+1)x^{k}$$ multiply by $x^3$ $$x^3(\frac{x}{1+x})'=\sum_{k=0}^{\infty}(-1)^k(k+1)x^{k+3}$$ $$(x^3(\frac{x}{1+x})')'=\sum_{k=0}^{\infty}(-1)^k(k+1)(k+3)x^{k+2}$$ divide by $x^2$ $$\frac{1}{x^2}(x^3(\frac{x}{1+x})')'=\sum_{k=0}^{\infty}(-1)^k(k+1)(k+3)x^{k}$$ $$\frac{(x+3)}{(x+1)^3}=\sum_{k=0}^{\infty}(-1)^k(k+1)(k+3)x^{k}$$
now let $x=\frac{1}{3}$