For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ $P(0)$ is true, as this just says: :$\dbinom 0 0^2 = 1 = \dbinom {2 \times 0} 0$ This holds by definition. === Basis for the Induction === $P(1)$ 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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{i \mathop = 0}^k \binom k i^2 = \binom {2 k} k$ Then we need to show: :$\displaystyle \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2 = \binom {2 \left({k + 1}\right)} {k + 1}$ === 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 \left({\binom k {i - 1} + \binom k i}\right)^2 + 1       | c = Pascal's Rule }} {{eqn | r = 1 + \sum_{i \mathop = 1}^k \left({\binom k {i - 1}^2 + \binom k i^2 + 2 \binom k {i - 1} \binom k i}\right) + 1       | c =  }} {{eqn | r = \left({\sum_{i \mathop = 0}^{k - 1} \binom k i^2 + 1}\right) + \left({1 + \sum_{i \mathop = 1}^k \binom k i^2}\right) + 2 \sum_{i \mathop = 1}^k \left({\binom k {i - 1} \binom k i}\right)       | c = Translation of Index Variable of Summation }} {{eqn | r = \left({\sum_{i \mathop = 0}^{k - 1} \binom k i^2 + \binom k k^2}\right) + \left({\binom k 0^2 + \sum_{i \mathop = 1}^k \binom k i^2}\right) + 2 \sum_{i \mathop = 1}^k \left({\binom k {i - 1} \binom k i}\right)       | 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 \left({\binom k {i-1} \binom k i}\right)       | c = }} {{eqn | r = \binom {2 k} k + \binom {2 k} k + 2 \sum_{i \mathop = 1}^k \left({\binom k {i - 1} \binom k i}\right)       | c = Induction Hypothesis }} {{end-eqn}} Now we look at $\displaystyle 2 \sum_{i \mathop = 1}^k \left({\binom k {i - 1} \binom k i}\right)$. 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: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2       | r = 2 \binom {2 k} k + 2 \sum_i \left({\binom k {i-1} \binom k i}\right)       | 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 \left({k + 1}\right)} {k + 1}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N: \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ {{qed}}