Well done Ang Zhi Ping, age 15, River Valley High School, Singapore for your excellent solution to this question.

Let the binomial coefficient $n! / r! (n - r)!$ be denoted by

$(\binom{n}{r}) .$ $(1)$

By considering powers of $(1 + x)$ show that

$\sum_{k = 0}^{n} {(\binom{n}{k})}^{2} = (\binom{2 n}{n})$

As $(1 + x)^{n} (1 + x)^{n} = (1 + x)^{2 n} (1)$ , we write down the Binomial expansion giving:

$[\sum_{p = 0}^{n} (\binom{n}{p}) x^{p}] [\sum_{q = 0}^{n} (\binom{n}{q}) x^{q}] = \sum_{r = 0}^{2 n} (\binom{2 n}{r}) x^{r} .$

The left hand side of the equation is

$[(\binom{n}{0}) + (\binom{n}{1}) x + \dots + (\binom{n}{n}) x^{n}] [(\binom{n}{0}) + (\binom{n}{1}) x + \dots + (\binom{n}{n}) x^{n}] .$

So the coefficient of $x^{n}$ on the left hand side of (1) is

$(\binom{n}{0}) (\binom{n}{n}) + (\binom{n}{1}) (\binom{n}{n - 1}) + (\binom{n}{2}) (\binom{n}{n - 2}) + \dots + (\binom{n}{n - 1}) (\binom{n}{1}) + (\binom{n}{n}) (\binom{n}{0}) .$

Since

$(\binom{n}{r}) = (\binom{n}{n - r})$ $(2)$
we see that the coefficient of $x^{n}$ on the left hand side of (1) is

$\sum_{k = 0}^{n} {(\binom{n}{k})}^{2} .$ $(3)$
As the coefficient of $x^{n}$ on the right hand side of (1) is

$(\binom{2 n}{n})$ $(4)$
the given formula is proven.