I want to prove the following identity using induction (not double counting method). Although it is a specific version of Vandermonde's identity and its inductive proof is presented here, but I need a direct inductive proof on this, not the general form.
$$\binom{2n}{n}=\sum\limits_{k=0}^n\binom{n}{k}^2$$
I have tried to simplify $\binom{2n+2}{n+1}$ using Pascal's theorem, but did not get any result. Any help?
Best Answer
Suppose $$ \sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}\tag{1} $$ then $$ \begin{align} \sum_{k=0}^{n+1}\binom{n+1}{k}^2 &=\sum_{k=0}^{n+1}\left[\binom{n}{k}+\binom{n}{k-1}\right]^2\tag{2a}\\ &=\sum_{k=0}^{n+1}\left[\binom{n}{k}^2+\binom{n}{k-1}^2+2\binom{n}{k}\binom{n}{k-1}\right]\tag{2b}\\ &=\binom{2n}{n}\left(1+1+\frac{2n}{n+1}\right)\tag{2c}\\ &=\binom{2n+2}{n+1}\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: Pascal's Triangle identity
$\text{(2b)}$: algebra
$\text{(2c)}$: apply $(1)$ and $(3)$
$\text{(2d)}$: $\binom{2n+2}{n+1}=\frac{4n+2}{n+1}\binom{2n}{n}$
Lemma: $$ \sum_{k=1}^n\binom{n}{k}\binom{n}{k-1}=\frac{n}{n+1}\sum_{k=0}^n\binom{n}{k}^2\tag{3} $$ Proof:
Since $\binom{n}{k-1}=\frac{k}{n-k+1}\binom{n}{k}$, we have $\binom{n}{k}+\binom{n}{k-1}=\frac{n+1}{n-k+1}\binom{n}{k}$. Therefore, $$ \frac{n-k+1}{n+1}\left[\binom{n}{k}+\binom{n}{k-1}\right]\binom{n}{k-1}=\binom{n}{k}\binom{n}{k-1}\tag{3a} $$ Since $\binom{n}{k}=\frac{n-k+1}{k}\binom{n}{k-1}$, we have $\binom{n}{k}+\binom{n}{k-1}=\frac{n+1}{k}\binom{n}{k-1}$. Therefore, $$ \frac{k}{n+1}\left[\binom{n}{k-1}+\binom{n}{k}\right]\binom{n}{k}=\binom{n}{k-1}\binom{n}{k}\tag{3b} $$ Adding $(3a)$ and $(3b)$ and cancelling yields $$ \frac{n-k+1}{n+1}\binom{n}{k-1}^2+\frac{k}{n+1}\binom{n}{k}^2=\binom{n}{k-1}\binom{n}{k}\tag{3c} $$ Summing $(3c)$ over $k$, and substituting $k\mapsto k+1$ in the leftmost sum, gives $$ \frac{n}{n+1}\sum_{k=0}^n\binom{n}{k}^2=\sum_{k=1}^n\binom{n}{k-1}\binom{n}{k}\tag{3d} $$ QED