Sum of Squares of Binomial Coefficients/Inductive Proof

Proof
For all $n \in \N$, let $\map P n$ be the proposition:
 * $\ds \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$

$\map P 0$ is true, as this just says:
 * $\dbinom 0 0^2 = 1 = \dbinom {2 \times 0} 0$

This holds by definition.

Basis for the Induction
$\map P 1$ is true, as this just says:
 * $\dbinom 1 0^2 + \dbinom 1 1^2 = 1^2 + 1^2 = 2 = \dbinom 2 1$

This also holds by definition.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:


 * $\ds \sum_{i \mathop = 0}^k \binom k i^2 = \binom {2 k} k$

Then we need to show:


 * $\ds \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2 = \binom {2 \paren {k + 1} } {k + 1}$

Induction Step
This is our induction step:

Now we look at $\ds 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}$.

Using the Chu-Vandermonde Identity:
 * $\ds \sum_i \binom r i \binom s {n - i} = \binom {r + s} n$

From the Symmetry Rule for Binomial Coefficients, this can be written:
 * $\ds \sum_i \binom r i \binom s {s - n + i} = \binom {r + s} n$

Putting $r = k, s = k, s - n = -1$ from whence $n = k + 1$:
 * $\ds \sum_i \binom k i \binom k {i - 1} = \binom {2 k} {k + 1}$

So:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \forall n \in \N: \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$