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

$$$$$$\mathrm{calculate}$$$$$$$$\mathrm{\Omega }=\underset{k=0}{\sum ^n}\frac{1}{(n-k)!.(n+k)!}$$$$$$

Answered by aleks041103 last updated on 12/Apr/23

$$\frac{1}{(n-k)!(n+k)!}=\frac{1}{\left(2n\right)!}\frac{\left(2n\right)!}{(n-k)!(2n-(n-k))!}=$$$$=\frac{1}{\left(2n\right)!}\left(\begin{array}{c}2n\\ n-k\end{array}\right)$$$$\Rightarrow \mathrm{\Omega }=\frac{1}{\left(2n\right)!}\underset{k=0}{\sum ^n}\left(\begin{array}{c}2n\\ n-k\end{array}\right)=$$$$=\frac{1}{\left(2n\right)!}\underset{k=0}{\sum ^n}\left(\begin{array}{c}2n\\ k\end{array}\right)=$$$$=\frac{1}{\left(2n\right)!}\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)+\underset{k=0}{\sum ^{2n}}\left(\begin{array}{c}2n\\ k\end{array}\right)}{2}=$$$$=\frac{2^{2n}+\frac{\left(2n\right)!}{{(n!)}^2}}{\left(2n\right)!}$$$$$$$$\Rightarrow \mathrm{\Omega }=\frac{4^n}{\left(2n\right)!}+\frac{1}{{(n!)}^2}$$

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