Find a simple expression for $\sum_{k=0}^{n} k(k-1)\binom{n}{k}$ for $n\geq0$ Find a simple expression for $\sum_{k=0}^{n} k(k-1)\binom{n}{k}$ for $n\geq0$.
My intuition is to somehow make use of the fact that $\binom{n}{k} = \frac{n}{k}\binom{n - 1}{k - 1}$. Can I simply replace the $\binom{n}{k}$ with this identity and solve from there somehow?

Let $X \sim \operatorname{Bin}(n,\tfrac{1}{2})$ be a binomial random variable. You are after $$2^n\mathbb{E}[X(X-1)] = 2^n(\mathbb{E}[X^2]-\mathbb{E}[X]) = 2^n(\mathbb{V}[X] + \mathbb{E}[X]^2 - \mathbb{E}[X]).$$ Since expectation and variance are additive, it is easy to calculate $\mathbb{E}[X] = \tfrac{n}{2}$ and $\mathbb{V}[X] = \tfrac{n}{4}$. Therefore
$$2^n\mathbb{E}[X(X-1)] = 2^n\left(\frac{n}{4} + \frac{n^2}{4} - \frac{n}{2}\right) = 2^{n-2}n(n-1).$$
