Revision as of 01:40, 15 June 2024 by Admin (Created page with "Prove that <math display="block"> u^{(2)}_{2n} = \frac 1{4^{2n}} \sum_{k = 0}^n \frac {(2n)!}{k!k!(n-k)!(n-k)!}\ , </math> and <math display="block"> 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 display="block"> \frac 1{2^{2n}}{{2n}\choose n} \sum_{...")
ABy Admin
Jun 15'24
Exercise
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]]