Proof of $\sum_{k=1}^n \binom{n}{k} \binom{n}{n+1-k} = \binom{2n}{n+1}$ via induction

Question

I'm having trouble to prove the following formula using Induction on $n \in \mathbb{N}$: $$\sum_{k=1}^n \binom{n}{k} \binom{n}{n+1-k} = \binom{2n}{n+1}.$$ I've tried all the usual identities, but they seem to lead nowhere. Is there any trick to this, or is it just not possible to prove this using induction?

I'm thankful for any tip or advice on how to approach this :)

If you search a bit in Approach0 you can find two problems which are basically modifications of this: Prove $\sum_{k=1}^{n}{n \choose k}{n \choose k-1} = {2n+2 \choose n+1}/2-{2n \choose n}$ and Evaluate the sum $\sum_{k=1}^n {n \choose k-1} {n \choose k}$. As mentioned in the previous comment, look at Vandermonde's identity. — Martin Sleziak, Dec 12 '17 at 17:04
However, neither of those two posts asks specifically for a proof by induction. — Martin Sleziak, Dec 12 '17 at 17:14
Then we could do induction on $i$, so the full thesis is just the base $P_0$ of the induction, and $P_i\implies P_{i+1}$ is easy because in fact $P_i= P_{i+1}$ for all $i$ — Pietro Majer, Dec 12 '17 at 17:35
Thanks for answering everyone. I know the formula is easily proven in various ways, but I was looking specifically for an Induction.
I had to settle with the an inductive proof of Vandermonde's identity though, obtaining this formula as a special case. — NumberPractice, Dec 12 '17 at 19:03
Maybe checking these related posts might help: Inductive proof for $\binom{2n}{n}=\sum\limits_{k=0}^n\binom{n}{k}^2$ and Inductive Proof for Vandermonde's Identity? Especially the first one, since robjohn's answer also deals with the sum $\sum_{k=1}^n\binom{n}{k}\binom{n}{k-1}$. — Martin Sleziak, Dec 13 '17 at 01:34
Neither Evaluate the sum $\sum_{k=1}^n {n \choose k-1} {n \choose k}$ nor any of the answers thereto concern induction. If this is to be closed, it would be nice to cite an inductive proof for those looking for an inductive proof. — robjohn, Dec 14 '17 at 11:18
What are you looking for beyond the Inductive Proof for Vandermonde's Identity? It seems like that can be easily adapted to prove what you want just by subtitutions of the variables. — rogerl, Dec 14 '17 at 16:45

score 3 · Answer 1 · edited Dec 12 '17 at 17:34

3

Select $n+1$ unit squares from a $2\times n$ rectangle. What do you see?

edited Dec 12 '17 at 17:34

Martin Sleziak

53,687

answered Dec 12 '17 at 16:22

Pietro Majer

275

1

Is that a hint to an inductive proof? – stochastic Dec 12 '17 at 16:40
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Actually you don't even need induction. The picture say it all. Select any $(n+1)$-subset of ${0,1}\times{1,\dots,n}$. You see a $k$-subset of the fist column and, independent of it, a $(n+1-k)$-subset of the other column. This exactly represents the LHS. – Pietro Majer Dec 12 '17 at 17:22
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Your suggestion leads to what I consider the "right" proof, the "natural" proof. (And it proves a more general result, about $\binom{2n}{m}$ for arbitrary $m$.) But the question specifically asked for a proof by induction on $n$. That seems to make the problem harder. – Andreas Blass Dec 12 '17 at 18:08

G Cab · Answer 2 · 2017-12-14T17:57:26.863

As already noted yours is just a Vandermonde convolution, which is normally proved by taking the product of two binomials.

However, if we limit to consider the general Vandermonde convolution with non-negative integer parameters, except at most one of the upper parametr which may also be real, i.e. $$ \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ m \cr k \cr} \right)\left( \matrix{ r \cr n - k \cr} \right)} = \left( \matrix{ r + m \cr n \cr} \right)\quad \left| \matrix{ \;0 \le m,n \in Z \hfill \cr \;r \in R \hfill \cr} \right. $$ then the convolution can be obtained as the $m$ times iteration of the basic recursive identity $$ \eqalign{ & \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ m \cr k \cr} \right)\left( \matrix{ r \cr n - k \cr} \right)} = \left( \matrix{ r + m \cr n \cr} \right)\quad \left| \matrix{ \;0 \le m,n \in Z \hfill \cr \;r \in R \hfill \cr} \right. \cr & \left( \matrix{ r + m \cr n \cr} \right) = \left( \matrix{ r + m - 1 \cr n \cr} \right) + \left( \matrix{ r + m - 1 \cr n - 1 \cr} \right) = \left( \matrix{ 1 \cr 0 \cr} \right)\left( \matrix{ r + m - 1 \cr n \cr} \right) + \left( \matrix{ 1 \cr 1 \cr} \right)\left( \matrix{ r + m - 1 \cr n - 1 \cr} \right) = \cr & = \left( \matrix{ r + m - 2 \cr n \cr} \right) + \left( \matrix{ r + m - 2 \cr n - 1 \cr} \right) + \left( \matrix{ r + m - 2 \cr n - 1 \cr} \right) + \left( \matrix{ r + m - 2 \cr n - 2 \cr} \right) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,2} \right)} {\left( \matrix{ 2 \cr k \cr} \right)\left( \matrix{ r + m - 2 \cr n - k \cr} \right)} = \cr & = \quad \cdots = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ m \cr k \cr} \right)\left( \matrix{ r \cr n - k \cr} \right)} \cr} $$

Now, the basic recursive identity can be taken as just the definition of the binomial , or if the binomial is defined in other ways it can be easily demonstrated also by induction.

And once you have demonstrated the basic recursion, its iteration leads you to demonstrate the Vandermonde convolution.

Proof of $\sum_{k=1}^n \binom{n}{k} \binom{n}{n+1-k} = \binom{2n}{n+1}$ via induction

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