Express it in its reduced form : $\sum\limits_{k=1}^{n}{(C(n,k-1)*C(n,k))}$ As we know $C^{2}(n,0)$+$C^{2}(n,1)$+$C^{2}(n,2)$+....+$C^{2}(n,n)$ =
$C (2n,n)$ 
By deducing it from $ (1+x)^{n}$
So, how can I find the reduced form of  $\sum\limits_{k=1}^{n}{(C(n,k-1)*C(n,k))}$
From  $ (1+x)^{n}$
Please help me to solve this,
Any help will be appreciated.

Let $[x^k]: (1+x)^n$ denote the coefficient of $x^k$ for the function $(1+x)^n$ ; that is $\binom{n}{k}$. Now
\begin{eqnarray*}
\sum_{k=1}^{n} \binom{n}{k-1} \binom{n}{k} = \sum_{k=1}^{n} [x^k]: \binom{n}{k-1}  (1+x)^n \\
=  [x^n]: \sum_{k=1}^{n}  \binom{n}{k-1} x^{n-k} (1+x)^n \\
= [x^n]:  x^{n-1}(1+\frac{1}{x})^n (1+x)^n \\
= [x^n]:  x^{-1}(1+x)^n (1+x)^n = \binom{2n}{n-1}.
\end{eqnarray*}
