# binomial identity $\sum\limits_{k\ge 0} (-1)^k{n+1\choose k}{n+m-4k\choose n}=\sum\limits_{k\ge 0}{n+1\choose k}{n\choose m-2k}$

Let $m,n$ be integers satisfying the condition $m\le 3(n+1)$.

Prove the identity $$\sum\limits_{k\ge 0} (-1)^k{n+1\choose k}{n+m-4k\choose n}=\sum\limits_{k\ge 0}{n+1\choose k}{n+1\choose m-2k}$$

I have just found the way solve this problem.  We have $$\left\langle x^m\right\rangle\left[(1+x+x^2+x^3)^{n+1}\right]=\left\langle x^m\right\rangle\left[(1+x)^{n+1}(1+x^2)^{n+1}\right]=\left\langle x^m\right\rangle\left[\left(\dfrac{1-x^4}{1-x}\right)^{n+1}\right]$$ $$\begin{eqnarray*}(1+x)^{n+1}(1+x^2)^{n+1}&=&\sum_{k=0}^{n+1}{n+1\choose k}x^k\sum_{j=0}^{n+1}{n+1\choose j}x^{2j}\\&=&\sum_{k=0}^{n+1}\sum_{j=0}^{n+1}{n+1\choose k}{n+1\choose j}x^{k+2j}\end{eqnarray*}$$ $$\tag{1}\Rightarrow \left\langle x^m\right\rangle\left[(1+x)^{n+1}(1+x^2)^{n+1}\right]=\sum_{j=0}^{n+1}{n+1\choose m-2j}{n+1\choose j}$$ Other way $$\begin{eqnarray*}\left(\dfrac{1-x^4}{1-x}\right)^{n+1}&=&\sum_{k=0}^{n+1}{n+1\choose k}(-1)^kx^{4k}\sum_{j\ge 0}{n+j\choose j}x^j\\&=&\sum_{j\ge 0}\sum_{k=0}^{n+1}{n+1\choose k}{n+j\choose j}(-1)^kx^{4k+j}\end{eqnarray*}$$ $$\tag{2}\Rightarrow \left\langle x^m\right\rangle\left[\left(\dfrac{1-x^4}{1-x}\right)^{n+1}\right]=\sum_{k\ge 0}{n+1\choose k}{n+m-4k\choose m-4k}(-1)^k$$ $(1)\quad\&\quad(2)\Rightarrow Q.E.D$