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?

  • 3
    $\begingroup$ You might not feel like adding $2020$ addends together by hand, but that does not make the series infinite :) $\endgroup$ – darij grinberg Oct 15 at 18:17
  • 1
    $\begingroup$ 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? $\endgroup$ – darij grinberg Oct 15 at 18:18
  • $\begingroup$ Sorry, I mean non-infinite series $\endgroup$ – Tomáš Macháček Oct 15 at 18:19
  • $\begingroup$ I know that $2^{2019}$ is same as $\sum_{n=0}^{2019} {2019 \choose n}$ $\endgroup$ – Tomáš Macháček Oct 15 at 18:21



$$(m-2n)^2=m^2+an+bn(n-1)$$ 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?

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

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