Show that $$\sum_{k=0}^{n}\binom{n}{k}\left(\binom{m+k+1}{m-n}+(-1)^{k+1}\binom{m+2(n-k)+1}{m}\right)=0$$ for all integers $m$ and $n$ with $m\ge n\ge 0$. I tried induction on $n$, but there's not a very nice way to change the LHS from the $n$ case to $n+1$.

  • 2
    $\begingroup$ Isn't the main problem with the induction transition that the different terms produce different results because they change at alternate ends of the range $k\in[0,n]$? Is it possible to split this into two sums to alleviate this issue? $\endgroup$
    – abiessu
    Nov 2, 2020 at 21:56

1 Answer 1


We get for the first of the two pieces

$$\sum_{k=0}^n {n\choose k} {m+k+1\choose m-n} = [z^{m-n}] (1+z)^{m+1} \sum_{k=0}^n {n\choose k} (1+z)^k \\ = [z^{m-n}] (1+z)^{m+1} (2+z)^n.$$

The second piece is

$$\sum_{k=0}^n {n\choose k} (-1)^{k+1} {m+2(n-k)+1\choose m} \\ = - [z^m] (1+z)^{m+2n+1} \sum_{k=0}^n {n\choose k} (-1)^k (1+z)^{-2k} \\ = - [z^m] (1+z)^{m+2n+1} \left(1-\frac{1}{(1+z)^2}\right)^n \\ = - [z^m] (1+z)^{m+1} (2z+z^2)^n = -[z^m] z^n (1+z)^{m+1} (2+z)^n \\ = -[z^{m-n}] (1+z)^{m+1} (2+z)^n.$$

This is the claim.


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