Well done Ang Zhi Ping from River Valley High School, Singapore for
your excellent solution to this question.
Let the binomial coefficient n!/r!(n-r)! be denoted by
\begin{equation}{n\choose r}.\end{equation}
By considering powers of (1 + x) show that \sum_{k=0}^n
{n\choose k}^2 = {2n \choose n}
As (1 + x)^n(1 + x)^n = (1 + x)^{2n} \quad (1) , we write down
the Binomial expansion giving: \begin{equation*}\left[\sum_{p=0}^n
{n\choose p} x^p\right]\left[\sum_{q=0}^n {n\choose q} x^q\right] =
\sum_{r=0}^{2n} {2n\choose r}x^r. \end{equation*}
The left hand
side of the equation is \begin{equation*}\left[{n\choose 0} +
{n\choose 1}x + \cdots + {n\choose n}x^n\right] \left[{n\choose 0}
+ {n\choose 1}x + \cdots + {n\choose n}x^n\right].\end{equation*}
So the coefficient of x^n on the left hand side of (1) is
\begin{equation*}{n\choose 0}{n\choose n} + {n\choose 1}{n\choose
n-1} + {n\choose 2}{n\choose n-2} + \cdots +{n\choose n-1}{n\choose
1} + {n\choose n}{n\choose 0}.\end{equation*}
Since
\begin{equation}{n\choose r} = {n\choose n-r}\end{equation}
we see
that the coefficient of x^n on the left hand side of (1) is
\begin{equation}\sum_{k=0}^n {n\choose k}^2.\end{equation}
As the
coefficient of x^n on the right hand side of (1) is
\begin{equation}{2n\choose n}\end{equation}