Calculate $\sum_{j=0}^k\binom {2k+1}{2j+1}^2=?$ Knowing that:
$${2k\choose k}=\sum_{j=0}^k{k\choose j}^2.$$
calculate the sums:
$$\sum_{j=0}^k\binom {2k+1}{2j+1}^2=?$$
Any sugestions please?
Thanks in advance.
 A: $$\sum_{j=0}^k{k\choose j}^2 = {2k\choose k}$$
$$\implies \sum_{j=0}^{2k+1}{2k+1\choose j}^2 = {4k+2\choose 2k+1} \ (replace \ k \ by \ 2k+1\ )$$
$$\implies \sum_{j=0}^{k}{2k+1\choose 2j+1}^2 + \sum_{j=0}^{k}{2k+1\choose 2j}^2 = {4k+2\choose 2k+1} \ (divide \ into \ even\ and\ odd \ size \ subsets)$$
$$\implies 2*\sum_{j=0}^{k}{2k+1\choose 2j+1}^2 = {4k+2\choose 2k+1} \ (since \ 2k+1 \ is\ odd \ and \ {2k+1\choose 2j+1} = {2k+1\choose 2k-2j})$$
$$\implies \sum_{j=0}^{k}{2k+1\choose 2j+1}^2 = \frac 1 2{4k+2\choose 2k+1}$$
A: Using the similar combinatorial arguments as for the first identity, we get
$$\sum_{j=0}^k\binom {2k+1}{2j+1}^2= \sum_{j=0}^k\binom {2k+1}{2j+1}\binom {2k+1}{2k -2j} = \dfrac{\binom {4k+2}{2k+1}}{2}$$
To pick $2k+1$ elements out of $4k+2$ elements, we can either pick an odd number of elements from the first $2k+1$ ones and an even number of elments from the last $2k+1$ ones, or pick an even number of elements from the first $2k+1$ ones and an odd number of elments from the last $2k+1$ ones. Both ways represent half of the total possibilities, i.e. $\frac{\binom {4k+2}{2k+1}}{2}$
Then to pick an odd number of elements from the first $2k+1$ ones and an even number of elments from the last ones such that in total we pick $2k+1$ elements, we have $\sum_{j=0}^k\binom {2k+1}{2j+1}\binom {2k+1}{2k -2j}$ ways, so we conclude the result.
Another way to say this is to simply remark that
\begin{align}2\sum_{j=0}^k\binom {2k+1}{2j+1}^2 &=\sum_{j=0}^k\binom {2k+1}{2j+1}\binom {2k+1}{2k -2j} + \sum_{j=0}^k\binom {2k+1}{2k -2j}\binom {2k+1}{2j+1} \\
&= \sum_{j=0}^{2k+1}\binom {2k+1}{j}\binom {2k+1}{2k+1 -j} \\
&= \sum_{j=0}^{2k+1}\binom {2k+1}{j}^2 =\binom {4k+2}{2k+1}
\end{align} 
The last equality is the given identity
