Why doesn't Vandermonde's identity work in this case?

Question

Given the identity here, to me it seems that this should be true: $$\sum_{i = 0}^{n}\binom{n+i} i\binom{2n-i}{n-i} = \binom{3n}{n}$$

But it seems that this is not the case.

It turns out that it is actually $$\sum_{i = 0}^{n}\binom{n+i} i\binom{2n-i-1}{n-i} = \binom{3n}{n}$$for $n\ge1$.

Why is this the case? Shouldn't the above technically follow Vandermonde's identity?

Answer 1

Here we have Chu-Vandermonde's identity in disguise. The crucial point is that we have the bound variable $i$ not only as lower index, but also as upper index of the binomial coeffcient. Nevertheless we can transform the binomial coefficient using the binomial identity \begin{align*} \binom{p+q-1}{q}=(-1)^q\binom{-p}{q}\tag{1} \end{align*} and then we can apply Chu-Vandermonde's identity.

We obtain \begin{align*} \color{blue}{\sum_{i = 0}^{n}}&\color{blue}{\binom{n+i} {i}\binom{2n-i-1}{n-i}}\\ &=\sum_{i=0}^n\binom{-n-1}{i}(-1)^i\binom{-n}{n-i}(-1)^{n-i}\tag{2}\\ &=(-1)^n\sum_{i=0}^n\binom{-n-1}{i}\binom{-n}{n-i}\tag{3}\\ &=(-1)^n\binom{-2n-1}{n}\tag{4}\\ &\,\,\color{blue}{=\binom{3n}{n}} \end{align*} and the claim follows.

Comment:

• In (2) we apply the binomial identity (1) twice.

• In (3) we are ready to apply the Chu-Vandermonde identity.

• In (4) we apply (1) again.