Can anybody help me to evaluate this sum $(1)$?
$$\sum_{j=0}^{k}\sum_{i=0}^{j}{k \choose j}^2{k \choose i}\tag1$$
I have manage to figure out:
$$\sum_{j=0}^{k}\sum_{i=0}^{j}{k \choose j}{k \choose i}=\frac{4^k+{2k \choose k}}{2}\tag2$$
and
$$\sum_{j=0}^{k}\sum_{i=0}^{j}{k+1 \choose j+1}^2{k \choose i}=2^k{2k \choose k}\frac{2k+1}{k+1}\tag3$$
Best Answer
Comment:
In (1) we use the coefficient of operator $[z^p]$ to denote the coefficient of $z^p$.
In (2) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
In (3) we select the coefficient of $z^k$.
Comment:
In (5) we change the order of the outer sum $j\to k-j$.
In (6) we shift the index $i$ by $k$.
In (7) we replace $i$ with $-i$.