Sum of Squares of Binomial Coefficients/Inductive Proof

Theorem

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

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

Proof

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

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

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

Basis for the Induction

$P(1)$ is true, as this just says $\displaystyle \binom 1 0^2 + \binom 1 1^2 = 1^2 + 1^2 = 2 = \binom 2 1$. 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:

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

Then we need to show:

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

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i=0}^{k+1} \binom {k+1} i^2\)	\(=\)	\(\displaystyle \binom {k+1} 0^2 + \sum_{i=1}^k \binom {k+1} i^2 + \binom {k+1} {k+1}^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \sum_{i=1}^k \left({\binom k {i-1} + \binom k i}\right)^2 + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from Pascal's Rule
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \sum_{i=1}^k \left({\binom k {i-1}^2 + \binom k i^2 + 2 \binom k {i-1} \binom k i}\right) + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\sum_{i=0}^{k-1} \binom k i^2 + 1}\right) + \left({1 + \sum_{i=1}^k \binom k i^2}\right) + 2 \sum_{i=1}^k \left({\binom k {i-1} \binom k i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Permutation of Indices
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\sum_{i=0}^{k-1} \binom k i^2 + \binom k k^2}\right) + \left({\binom k 0^2 + \sum_{i=1}^k \binom k i^2}\right) + 2 \sum_{i=1}^k \left({\binom k {i-1} \binom k i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=0}^k \binom k i^2 + \sum_{i=0}^k \binom k i^2 + 2 \sum_{i=1}^k \left({\binom k {i-1} \binom k i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {2 k} k + \binom {2 k} k + 2 \sum_{i=1}^k \left({\binom k {i-1} \binom k i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the induction hypothesis

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

Using the Chu-Vandermonde Identity:

$\displaystyle \sum_i \binom r i \binom s {n-i} = \binom {r+s} n$

From the Symmetry Rule for Binomial Coefficients, this can be written:

$\displaystyle \sum_i \binom r i \binom s {s - n + i} = \binom {r+s} n$

Putting $r = k, s = k, s - n = -1$ from whence $n = k + 1$:

$\displaystyle \sum_i \binom k i \binom k {i - 1} = \binom {2 k} {k+1}$

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i=0}^{k+1} \binom {k+1} i^2\)	\(=\)	\(\displaystyle 2 \binom {2 k} k + 2 \sum_i \left({\binom k {i-1} \binom k i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	when $i \le 0$ and $i > k$ we have $\displaystyle \binom k {i-1} \binom k i = 0$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 2 \binom {2 k} k + 2 \binom {2 k} {k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 2 \binom {2 k + 1} {k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Pascal's Rule
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {2 k + 1} k + \binom {2 k + 1} {k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Symmetry Rule for Binomial Coefficients
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {2 k + 2} {k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Pascal's Rule
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {2 \left({k + 1}\right)} {k+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

Therefore:

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

$\blacksquare$