Pensieve header: A combinatorial riddle from Assaf: Prove $\sum_{k=0}^n \binom{2(n-k)}{n-k}\binom{2k}{k}=4^n$.

AS[n_] := Sum[Binomial[2(n-k),n-k]Binomial[2k,k],{k,0,n}]

AS/@ Range[8]

DSolve[x f'[x]==4x f[x]+4(x^2) f'[x]-2x f[x]∧f[0]==1,f[x],x]