What is $\sum_{k=1}^n{{n}\choose{k}}k^{3}$ equal to?

Question

Find a compact form for the sum $$\sum_{k=1}^n{{n}\choose{k}}k^{3}$$

I have been trying to solve this question for some time now. My guess is that I am supposed to write this in terms of something simpler (like a power of a simple sum) but I just can't come up with anything. I would appreciate it if you could give me a hint before providing the full answer so I can think about it on my own.

Answer 1

Hint: Consider the simpler sum

$$\sum_{k=1}^n {n \choose k} kx^k = x\sum_{k=0}^n {n \choose k} kx^{k-1} = x\left(\sum_{k=0}^n {n \choose k} x^k\right)' = x((1+x)^n)'$$

(The sums can start at $0$ or $1$ since the $0$th term is $0$, but start at $0$ to make life easier). Now how would you do the $k^3$ case instead?

Answer 2

Hint: Prove the following

1.$\sum { n \choose k } k = n \sum { n - 1 \choose k - 1 } $.
2. $ \sum { n \choose k } k (k-1) = n (n-1) \sum { n - 2 \choose k - 2 }$.
3. A similar identity.