Compute $\sum\limits_{k=0}^{100}\frac{1}{(100-k)!(100+k)!}$ $$\sum_{k=0}^{100}\frac{1}{(100-k)!(100+k)!}$$
My work 
$$\sum_{k=0}^{100}\frac{2n!}{(2n!)(n-k)!(n+k)!}$$
$$\sum_{k=0}^{100}\frac{^{2n}C_{n-k}}{(2n!)}$$
$$\sum_{k=0}^{100}\frac{^{2n}C_{n+k}}{(2n!)}$$
How I should proceed further ?

Solving the problem for $n \in \mathbb{N}$ instead of only $n = 100$, you get
\begin{align}
\sum_{k=0}^{n} \frac{1}{(n-k)!(n+k)!}
&= \frac{1}{(2n)!} \sum_{k=n}^{2n} \frac{(2n)!}{k!(2n-k)!}
= \frac{1}{(2n)!} \sum_{k=n}^{2n} \binom{2n}{k}  \\
&= \frac{1}{2(2n)!} \left( \binom{2n}{n} + 2^{2n} \right) \\
\end{align}
In the last step you use the fact that
$$
\sum_{k=n}^{2n} \binom{2n}{k} = \frac{1}{2} \left( \binom{2n}{n} + 2^{2n} \right)
$$
which is shown by
\begin{align}
2^{2n} 
&= \sum_{k=0}^{2n} \binom{2n}{k} \\
&= \sum_{k=0}^{n-1} \binom{2n}{k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k}  \\
&= \sum_{k=n+1}^{2n} \binom{2n}{2n-k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k}  \\
&= \sum_{k=n+1}^{2n} \binom{2n}{k} ~+~ \binom{2n}{n} ~+~ \sum_{k=n+1}^{2n} \binom{2n}{k}  \\
&= 2 \sum_{k=n}^{2n} \binom{2n}{k} ~-~ \binom{2n}{n}
\end{align}
from which you get
$$
\frac{1}{2} \left( 2^{2n} + \binom{2n}{n} \right)
= \sum_{k=n}^{2n} \binom{2n}{2n-k} 
$$
