calculate-k-0-n-1-n-k-n-k-

Question Number 190788 by mnjuly1970 last updated on 11/Apr/23

calculate

Ω = \underset{k = 0}{\sum^{n}} \frac{1}{(n - k)! . (n + k)!}

Answered by aleks041103 last updated on 12/Apr/23

\frac{1}{(n - k)! (n + k)!} = \frac{1}{(2 n)!} \frac{(2 n)!}{(n - k)! (2 n - (n - k))!} =

= \frac{1}{(2 n)!} (\begin{matrix} 2 n \\ n - k \end{matrix})

\Rightarrow Ω = \frac{1}{(2 n)!} \underset{k = 0}{\sum^{n}} (\begin{matrix} 2 n \\ n - k \end{matrix}) =

= \frac{1}{(2 n)!} \underset{k = 0}{\sum^{n}} (\begin{matrix} 2 n \\ k \end{matrix}) =

= \frac{1}{(2 n)!} \frac{(\begin{matrix} 2 n \\ n \end{matrix}) + \underset{k = 0}{\sum^{2 n}} (\begin{matrix} 2 n \\ k \end{matrix})}{2} =

= \frac{2^{2 n} + \frac{(2 n)!}{{(n!)}^{2}}}{(2 n)!}

\Rightarrow Ω = \frac{4^{n}}{(2 n)!} + \frac{1}{{(n!)}^{2}}