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Question Number 142023 by SOMEDAVONG last updated on 25/May/21

$${\mathrm{S}}_{\mathrm{n}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{1}{(\mathrm{n}-\mathrm{k})!(\mathrm{n}+\mathrm{k})!}=??$$

Answered by aleks041103 last updated on 25/May/21

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

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