Prove that

[[math]] u^{(2)}_{2n} = \frac 1{4^{2n}} \sum_{k = 0}^n \frac {(2n)!}{k!k!(n-k)!(n-k)!}\ , [[/math]]

and

[[math]] u^{(3)}_{2n} = \frac 1{6^{2n}} \sum_{j,k} \frac {(2n)!}{j!j!k!k!(n-j-k)!(n-j-k)!}\ , [[/math]]

where the last sum extends over all non-negative [math]j[/math] and [math]k[/math] with [math]j+k \le n[/math]. Also show that this last expression may be rewritten as

[[math]] \frac 1{2^{2n}}{{2n}\choose n} \sum_{j,k} \biggl(\frac 1{3^n}\frac{n!}{j!k!(n-j-k)!}\biggr)^2\ . [[/math]]