Sum of Squares of Binomial Coefficients

Theorem

 * $$\sum_{i=0}^n \binom n i^2 = \binom {2 n} n$$

where $$\binom n i$$ is a binomial coefficient.

Proof
For all $$n \in \N$$, let $$P \left({n}\right)$$ be the proposition:
 * $$\sum_{i=0}^n \binom n i^2 = \binom {2 n} n$$

$$P(0)$$ is true, as this just says $$\binom 0 0^2 = 1 = \binom {2 \times 0} 0$$. This holds by definition.

Basis for the Induction
$$P(1)$$ is true, as this just says $$\binom 1 0^2 + \binom 1 1^2 = 1^2 + 1^2 = 2 = \binom 2 1$$. This also holds by definition.

This is our basis for the induction.

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

So this is our induction hypothesis:


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

Then we need to show:


 * $$\sum_{i=0}^{k+1} \binom {k+1} {i}^2 = \binom {2 \left({k+1}\right)} {k+1}$$

Induction Step]
This is our induction step:

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Now we look at $$2 \sum_{i=1}^{k} \left({\binom {k} {i-1} \binom {k} {i}}\right)$$.

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

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

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

So:

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Finally we note:

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

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