# Finite series with binomial coefficient

There is a finite series:

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

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

• You might not feel like adding $2020$ addends together by hand, but that does not make the series infinite :) – darij grinberg Oct 15 at 18:17
• What have you tried? Of course the first step is to replace $2019$ by $m$. Now, if you expand the square, do you get anything more familiar? – darij grinberg Oct 15 at 18:18
• Sorry, I mean non-infinite series – Tomáš Macháček Oct 15 at 18:19
• I know that $2^{2019}$ is same as $\sum_{n=0}^{2019} {2019 \choose n}$ – Tomáš Macháček Oct 15 at 18:21

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?

• I am not sure that I understand this solution (hint), could you explain it. – Tomáš Macháček Oct 15 at 19:22
• I found that $\sum_{0}^{m} \frac{(m-2n)^2}{2^m- { m \choose n}} = m$ – Tomáš Macháček Oct 15 at 19:44
• @TomášMacháček, Please find the updated answer – lab bhattacharjee Oct 16 at 6:04