How to prove that $\sum_{k=0}^n \binom{n}{k} z^k(1-z)^{n-k} (k-nz)^2 = nz(1-z)$?

Question

for a practice exercise I have to prove that $\sum_{k=0}^n \binom{n}{k} z^k(1-z)^{n-k} (k-nz)^2 = nz(1-z)$ using the binomial theorem, but I can't work it out. I know obviously that $1 = (z+(1-z))^n = \sum_{k=0}^n \binom{n}{k} z^k(1-z)^{n-k}$ but I don't know how to work the $(k-nz)^2$ in there.

Answer 1

Here is a proof of a more general identity.

Let $n$ be a non-negative integer, and consider the identity

$$ \sum_{k=0}^{\infty} \binom{n}{k}z^k w^{n-k} e^{(kw - (n-k)z)s} = (z e^{ws} + w e^{-zs})^n $$

in three variables $z$, $w$, and $s$, which is a consequence of the binomial theorem. Differentiating both sides with respect $s$, we get

$$ \sum_{k=0}^{\infty} \binom{n}{k}z^k w^{n-k} (kw - (n-k)z) e^{(kw - (n-k)z)s} = n zw (e^{ws} - e^{-zs}) (z e^{ws} + w e^{-zs})^{n-1}. $$

Differentiating both sides with respect $s$ again,

\begin{align*} &\sum_{k=0}^{\infty} \binom{n}{k}z^k w^{n-k} (kw - (n-k)z)^2 e^{(kw - (n-k)z)s} \\ &\quad = n zw (w e^{ws} + z e^{-zs}) (z e^{ws} + w e^{-zs})^{n-1} \\ &\qquad + n(n-1) z^2w^2 (e^{ws} - e^{-zs})^2 (z e^{ws} + w e^{-zs})^{n-2}. \end{align*}

Plugging $s = 0$ to both sides, we get

$$ \sum_{k=0}^{\infty} \binom{n}{k}z^k w^{n-k} (kw - (n-k)z)^2 = n zw (z + w)^n. $$

If $w = 1-z$, then this identity reduces to the one asked by OP.