I was trying to prove some equation and reached this summation $\sum \limits_{j=0}^{m} {2j\choose{j}} {2m-2j \choose {m-j}} = 4^m $
I tried pascal identity and other known binomial identities but with no succes
Is there any simple proof (preferably algebraic one).
Thanks
Note that for any real $ \alpha $ : $$ \left(\forall x\in\left]-1,1\right[\right),\ \displaystyle\sum_{n=0}^{+\infty}{\displaystyle\binom{\alpha}{n}x^{n}}=\left(1+x\right)^{\alpha} $$ Thus, for $ x\in\left]-1,1\right[ $ : $$ \displaystyle\sum_{n=0}^{+\infty}{\left(-1\right)^{n}\displaystyle\binom{-\frac{1}{2}}{n}x^{n}}=\displaystyle\frac{1}{\sqrt{1-x}} \ \ \ \ \left(*\right) $$ Observe that : $$ \left(\forall n\in\mathbb{N}\right),\ \left(-1\right)^{n}\displaystyle\binom{-\frac{1}{2}}{n}=\displaystyle\frac{\left(-1\right)^{n}}{n!}\displaystyle\prod_{k=0}^{n-1}{\left(-\displaystyle\frac{1}{2}-k\right)}=\displaystyle\frac{1}{2^{n}n!}\displaystyle\prod_{k=0}^{n-1}{\left(2k+1\right)}=\displaystyle\frac{1}{4^{n}}\displaystyle\binom{2n}{n} $$ Multiplying the expression $ \left(*\right) $ by itself gives the following : $$ \displaystyle\sum_{n=0}^{+\infty}{\displaystyle\frac{1}{4^{n}}\left(\displaystyle\sum_{k=0}^{n}{\displaystyle\binom{2k}{k}\displaystyle\binom{2n-2k}{n-k}}\right)x^{n}}=\displaystyle\frac{1}{1-x}=\displaystyle\sum_{n=0}^{+\infty}{x^{n}} $$ Then we get the result : $$ \left(\forall n\in\mathbb{N}\right),\ \displaystyle\sum_{k=0}^{n}{\displaystyle\binom{2k}{k}\displaystyle\binom{2n-2k}{n-k}}=4^{n} $$
