I'm asked to prove the identity $$n(n+1)2^{n-2} = \sum_{k=1}^{n}k^2\binom{n}{k}$$ using the binomial theorem.
What I've been able to come up with so far is letting $$f(x) = (x+1)^n = \sum_{k=0}^{n}x^k\binom{n}{k}$$ taking two derivatives gives $$f''(x) = n(n-1)(x+1)^{n-2} = \sum_{k=2}^{n}k(k-1)x^{k-2}\binom{n}{k}$$ If I plug in x = 1, I get something super close to the required identity: $$n(n-1)2^{n-2} = \sum_{k=2}^{n}k(k-1)\binom{n}{k}$$
I think the method I'm using is correct, and I just need to tweak it a little bit... Any help is appreciated!
After you take the first derivative, you have
$$n(x+1)^{n-1}=\sum_{k=1}^nkx^{k-1}\binom{n}k\;.$$
Now multiply both sides by $x$ before you differentiate again.
Here’s an alternative that uses the binomial theorem only to evaluate $\sum_k\binom{m}k=2^m$. Using the fact that $k\binom{n}k=n\binom{n-1}{k-1}$, we can calculate
$$\begin{align*} \sum_kk^2\binom{n}k&=\sum_kkn\binom{n-1}{k-1}\\ &=n\sum_kk\binom{n-1}{k-1}\\ &=n\sum_k(k+1)\binom{n-1}k\\ &=n\sum_kk\binom{n-1}k+n\sum_k\binom{n-1}k\\ &=n\sum_k(n-1)\binom{n-2}{k-1}+n2^{n-1}\;, \end{align*}$$
and I expect that you can finish it from there.