Combinatorial coefficients squared If $C_0,C_1,C_2,...,C_n$ are the combinatorial coefficients in the expansion of $(1+x)^n$ then prove that:
$$1C_0^2+3C_1^2+5C_3^2+...+(2n+1)C_n^2=\dfrac{(n+1)(2n)!}{n!n!}=(n+1)\binom {2n}n$$
I am able to compute the linear addition but not the squares of coefficients.
Thanks!

Note that 
$C_r=\binom nr$. 
It is a well-known result that $\sum_{r=0}^n\binom nr^2=\binom {2n}n$ which can be proven easily using the Vandermonde identity. 
For even $n$:
The number elements is odd, index from $0$ to $(\frac n2-1)$, $n$, and from $(\frac n2+1)$ to $n$
Since $\binom nr=\binom n{n-r}$, then $\sum_{r=0}^{\frac n2-1}\binom nr^2=\frac 12 \binom {2n}n$.
The summation in the question can then be written as
$$\begin{aligned}
&\begin{array}r
1\binom n0^2
&+3\binom n1 ^2
&+\cdots
&+(n-1)\binom n{\frac n2-1}^2\\
& & & & +(n+1) \binom n{\frac n2} ^2\\
+(2n+1)\binom nn^2
&+(2n-1)\binom n{n-1}^2
&+\cdots
&+(n+3)\binom n{\frac n2+1}^2
\\
=(2n+2)\binom n0^2
&+(2n+2)\binom n1^2
&+\cdots
&+(2n+2)\binom n{\frac n2-1}^2
&+(n+1)\binom n{\frac n2}^2
\\
\end{array}\\
&=(2n+2)\cdot \displaystyle\sum_{r=0}^{\frac n2-1}\binom nr^2+(n+1)\binom n{\frac n2}\\
&=(2n+2)\cdot \frac 12 \left[\displaystyle\sum_{r=0}^{n}\binom nr^2-\binom n{\frac n2}\right]+(n+1)\binom n{\frac n2}\\
&=(n+1)\left[\binom {2n}n-\binom n{\frac n2}\right]+(n+1)\binom n{\frac n2}\\
&=\color{red}{(n+1)\binom {2n}n}
\end{aligned}$$
__
For odd $n$:
The number elements is even, index from $0$ to $\frac {n-1}2$, and from $\frac {n+1}2$ to $n$
Since $\binom nr=\binom n{n-r}$, then $\sum_{r=0}^\frac {n-1}2 \binom nr^2=\frac 12 \binom {2n}n$.
The summation in the question can then be written as
$$\begin{aligned}
&\begin{array}r
1\binom n0^2
&+3\binom n1 ^2
&+\cdots
&+n\binom n{\frac {n-1}2}^2\\
+(2n+1)\binom nn^2
&+(2n-1)\binom n{n-1}^2
&+\cdots
&+(n+2)\binom n{\frac {n+1}2}^2
\\
=(2n+2)\binom n0^2
&+(2n+2)\binom n1^2
&+\cdots
&+(2n+2)\binom n{\frac{n-1}2}^2\\
\end{array}\\
&=(2n+2)\displaystyle\sum_{r=0}^{\frac {n-1}2}\binom nr^2\\
&=(2n+2)\cdot \frac 12\displaystyle\sum_{r=0}^{n}\binom nr^2\\
&=(2n+2)\cdot \frac 12 \binom {2n}n\\
&=\color{red}{(n+1)\binom {2n}n}
\end{aligned}$$
