Prove an equation about binomial coefficients

Question

Could we prove:

$ \sum_{k} \binom{2k}{k}\binom{n+k}{m+2k} \frac{(-1)^k}{k+1} = \binom{n-1}{m-1}$ when $m,n \in N$

Answer 1

When $n=1$ it's easy to evaluate for all $m\ge1$. Then it's easy to prove the $n+1$ by induction, given the $n$ case (for all $m$): just write $$ \binom{n+k}{m+2k} = \binom{n-1+k}{m+2k} + \binom{n-1+k}{m-1+2k}. $$

Answer 2

The generating function for the Central Binomial Coefficients is $$ \sum_{k=0}^\infty\binom{2k}{k}x^k=(1-4x)^{-1/2}\tag{1} $$ Integrating $(1)$ and dividing by $x$ yields $$ \sum_{k=0}^\infty\binom{2k}{k}\frac{x^k}{k+1}=\frac{1-\sqrt{1-4x}}{2x}\tag{2} $$ Sum the formula against $x^m$: $$ \begin{align} &\sum_{m=-\infty}^n\left(\sum_{k=0}^{n-m}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}\right)x^m\tag{3}\\ &=\sum_{k=0}^\infty\sum_{m=-2k}^{n-k}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}x^m\tag{4}\\ &=\sum_{k=0}^\infty x^{-2k}\sum_{m=0}^{n+k}\binom{2k}{k}\binom{n+k}{m}\frac{(-1)^k}{k+1}x^m\tag{5}\\ &=\sum_{k=0}^\infty x^{-2k}\binom{2k}{k}\frac{(-1)^k}{k+1}(x+1)^{n+k}\tag{6}\\ &=(x+1)^n\sum_{k=0}^\infty\binom{2k}{k}\frac{(-1)^k}{k+1}\left(\frac{x+1}{x^2}\right)^{k}\tag{7}\\ &=(x+1)^n\frac{-1+\sqrt{1+4\frac{x+1}{x^2}}}{2\frac{x+1}{x^2}}\tag{8}\\ &=x(x+1)^{n-1}\frac{-x+(x+2)}{2}\tag{9}\\[9pt] &=x(x+1)^{n-1}\tag{10}\\[6pt] &=\sum_{m=1}^n\binom{n-1}{m-1}x^m\tag{11} \end{align} $$ Justification:
$\:\ (3)$: sum against $x^m$
$\:\ (4)$: change order of summation
$\:\ (5)$: substitute $m\mapsto m-2k$
$\:\ (6)$: apply the Binomial Theorem
$\:\ (7)$: reorganize
$\:\ (8)$: apply $(2)$
$\:\ (9)$: simplify
$(10)$: simplify
$(11)$: apply the Binomial Theorem

Equating coefficients of $x^m$ yields $$ \sum_{k=0}^{n-m}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}=\binom{n-1}{m-1}\tag{12} $$

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Note that the sum has non-zero terms for $m\lt1$, but these cancel; e.g. $n=3,m=-1$: $$ \begin{align} &\binom{0}{0}\binom{3}{-1}\frac11-\binom{2}{1}\binom{4}{1}\frac12+\binom{4}{2}\binom{5}{3}\frac13-\binom{6}{3}\binom{6}{5}\frac14+\binom{8}{4}\binom{7}{7}\frac15\\[6pt] &=0-4+20-30+14\\[9pt] &=0 \end{align} $$