Simplify: $\sum_{k=1}^n ({n \choose k}k^2 )$

Question

I need to simplify: $\sum_{k=1}^n ({n \choose k}k^2 )$. My first idea was to get rid of the square and use known identities to proceed. This is what I did
$$\sum_{k=1}^n( {n \choose k}k^2) = \sum_{k=1}^n( \frac n k {n-1 \choose k-1}k^2) = \sum_{k=1}^n {n-1 \choose k-1} nk = n \sum_{k=1}^n {n-1 \choose k-1} k = \\ n \sum_{s=0}^n \left({n-1 \choose s}(s+1) \right)$$ Now, the only thing that stops me here is the $s+1$ factor. Is there a way to dispose of it, or maybe to change it to $s$?

Answer 1

Hint

You reached $$n \sum_{s=0}^{n-1} {n-1 \choose s}(s+1)$$ Let's continue $$n \sum_{s=0}^{n-1} {n-1 \choose s}(s+1) = n \left[\sum_{s=0}^{n-1} {n-1 \choose s}(s)+ \sum_{s=0}^{n-1} {n-1 \choose s}\right]$$ $$=n \left[(n-1)\sum_{s=1}^{n-1} {n-2 \choose {s-1}}+ \sum_{s=0}^{n-1} {n-1 \choose s} \right]$$ By binomial theorem

$$=n[(n-1)2^{n-2}+2^{n-1}]$$ $$=n(n+1).2^{n-2}$$

Answer 2

Hint : $$\sum_{k=1}^n \binom{n}{k}k^2 =\sum_{k=1}^n \binom{n}{k}k(k-1)+ \sum_{k=1}^n \binom{n}{k}k$$ And if you write $$ P(x)= \sum_{k=1}^n \binom{n}{k}x^k= (1+x)^n -1$$ You will have : $$\sum_{k=1}^n \binom{n}{k}k^2 =P''(1)+P'(1)$$