Theorem: Sum of Squares of Binomial Coefficients

{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \binom n i^2
      | r = \binom {2 n} n
}}
{{eqn | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}
}}
{{end-eqn}}

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:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2
      | r = \binom {k + 1} 0^2 + \sum_{i \mathop = 1}^k \binom {k + 1} i^2 + \binom {k + 1} {k + 1}^2
      | c =
}}
{{eqn | r = 1 + \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} + \binom k i}^2 + 1
      | c = Pascal's Rule
}}
{{eqn | r = 1 + \sum_{i \mathop = 1}^k \paren {\binom k {i - 1}^2 + \binom k i^2 + 2 \binom k {i - 1} \binom k i} + 1
      | c = 
}}
{{eqn | r = \paren {\sum_{i \mathop = 0}^{k - 1} \binom k i^2 + 1} + \paren {1 + \sum_{i \mathop = 1}^k \binom k i^2} + 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = \paren {\sum_{i \mathop = 0}^{k - 1} \binom k i^2 + \binom k k^2} + \paren {\binom k 0^2 + \sum_{i \mathop = 1}^k \binom k i^2} + 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}
      | c =
}}
{{eqn | r = \sum_{i \mathop = 0}^k \binom k i^2 + \sum_{i \mathop = 0}^k \binom k i^2 + 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}
      | c =
}}
{{eqn | r = \binom {2 k} k + \binom {2 k} k + 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}
      | c = Induction Hypothesis
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2
      | r = 2 \binom {2 k} k + 2 \sum_i \paren {\binom k {i - 1} \binom k i}
      | c = because when $i \le 0$ and $i > k$ we have $\dbinom k {i - 1} \dbinom k i = 0$
}}
{{eqn | r = 2 \binom {2 k} k + 2 \binom {2 k} {k + 1}
      | c = 
}}
{{eqn | r = 2 \binom {2 k + 1} {k + 1}
      | c = Pascal's Rule
}}
{{eqn | r = \binom {2 k + 1} k + \binom {2 k + 1} {k + 1}
      | c = Symmetry Rule for Binomial Coefficients
}}
{{eqn | r = \binom {2 k + 2} {k + 1}
      | c = Pascal's Rule
}}
{{eqn | r = \binom {2 \paren {k + 1} } {k + 1}
      | c = 
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
{{qed}}