Finite series with binomial coefficient

Question

There is a finite series:

\begin{equation} \mathscr{S} = \sum_{n=0}^{m} {m \choose n}(m-2n)^2 \end{equation}

If I divide $\mathscr{S}$ by $2^{m}$ I get m. Could anyone explain it to me?

Answer 1

Hint

Write

$$(m-2n)^2=m^2+an+bn(n-1)$$ where $a,b$ are arbitrary constants

so that $$\binom mn(m-2n)^2=m^2\binom mn+a n\cdot\binom mn+bn(n-1)\cdot\binom mn$$

Can you find $a,b?$

For $n\ge1,$ $$n\binom mn=\cdots=m\binom{m-1}{n-1}$$

For $n\ge2,$ $$n(n-1)\binom mn=\cdots=m(m-1)\binom{m-2}{n-2}$$

Can you take it from here?