$$\sum_{k=5}^\infty{{k-1}\choose{k-5}}\frac{k^3}{2^k}$$
I can't seem to simplify this sum. I get to a certain point then I get stuck, I know there must be some sort of trick to simplify it but I am unsure of how to do so. Any hints would be more than helpful!
Your sum equals: $$\sum_{j=0}^{+\infty}\frac{(j+5)^3}{32\cdot 2^j}\binom{j+4}{j}.\tag{1}$$ Since: $$\sum_{j=0}^{+\infty}\binom{j+4}{j}x^j = \frac{1}{(1-x)^5},\tag{2}$$ by differentiating three times both terms of $(2)$, you get the sums $\sum_{j\geq 0}\binom{j+4}{4}j^k\,x^j$ with $k\in\{0,1,2,3\}$, hence recombining these pieces you get: $$\sum_{j=0}^{+\infty}\frac{(j+5)^3}{32\cdot 2^j}\binom{j+4}{j}=1330.$$