Show that $$ \sum_{k\ +\ j\ =\ n\atop{\vphantom{\LARGE A}0\ \le\ k,\phantom{A} j\ \le\ n}}{2k \choose k}{2j \choose j} = 4^{n} $$
I think use integral solve it. But I don't it,and this problem is from this.
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Noting that $\sum_{k=0}^\infty\binom{2k}{k}x^k=\frac1{\sqrt{1-4x}},$ $$\begin{align}\sum_{a_1+a_2+\cdots+a_k=n,0\le a_1,a_2,\cdots,a_k\le n}\binom{2a_1}{a_1}\binom{2a_2}{a_2}\cdots\binom{2a_k}{a_k}\qquad(\star)\end{align}$$ is the coefficient of $x^n$ when we use power series expansion for $$\left(\frac1{\sqrt{1-4x}}\right)^k=(1-4x)^{-\frac 12 k}.$$
Hence, noting that $$(1-4x)^{-\frac 12 k}=\sum_{n=0}^\infty\binom{-\frac 12k}{n}(-4x)^{n},$$ we get $$(\star)=\binom{-\frac 12 k}{n}(-4)^n=\frac{\left(-\frac 12 k\right)\left(-\frac 12 k-1\right)\left(-\frac 12 k-n+1\right)}{n!}\cdot (-4)^n=\frac{\left(\frac 12 k\right)\left(\frac 12 k+1\right)\left(\frac 12 k+n-1\right)}{n!}\cdot 4^n.$$
This can be represented as the following for $m\in\mathbb N$ :
$$(\star)=\binom{m+n-1}{n}\cdot 4^n$$ when $k=2m.$
$$(\star)=\frac{\binom{2n}{n}\cdot\binom{2m+2n-2}{2n}}{\binom{m+n-1}{n}}$$ when $k=2m-1.$
Your question is the $k=2$ case.
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