91影视

$\underset{\mathbf{k}\mathbf{=}\mathbf{0}}{\overset{\mathbf{鈭�}}{\mathbf{鈭�}}}\frac{\mathbf{k}\mathbf{!}}{\mathbf{\left(}\mathbf{2}\mathbf{k}\mathbf{\right)}\mathbf{!}}$

The first term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is obtained by substituting $k=0$in $\frac{k!}{\left(2k\right)!}$. Therefore, the value at $k=0$is:

$\frac{k!}{\left(2k\right)!}=\frac{0!}{(2脳0)!}$(Substituting)

$=\frac{0!}{0!}$

=1( Because 0!=1)

The first term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is 1 .

The second term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is obtained by substituting $k=1$in $\frac{k!}{\left(2k\right)!}$. Therefore, the value at $k=1$is:

$\frac{k!}{\left(2k\right)!}=\frac{1!}{(2脳1)!}$(Substituting)

$=\frac{1!}{2!}=\frac{1}{2}$

The sccond term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is$\frac{1}{2}$.

The third term of the series$\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is obtained by substituting $k=2$in $\frac{k!}{\left(2k\right)!}$. Therefore, the value at $k=2$is:

$\frac{k!}{\left(2k\right)!}=\frac{3!}{(2脳3)!}$(Substituting)

$=\frac{3!}{6!}=\frac{3!}{6脳5脳4脳3!}=\frac{1}{120}$

The third term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is $\frac{1}{120}$.

The fourth term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is obtained by substituting $k=3$in$\frac{k!}{\left(2k\right)!}$. Therefore, the value at $k=3$is:

$\frac{k!}{\left(2k\right)!}=\frac{4!}{(2脳4)!}$(Substituting)

$=\frac{4!}{8!}=\frac{4!}{8脳7脳6脳5脳4!}=\frac{1}{1680}$

The fourth term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is $\frac{1}{1680}$.

The fifth term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is obtained by substituting $k=4$in $\frac{k!}{\left(2k\right)!}$. Therefore, the value at $k=4$is:

$$\begin{gathered} \frac{k!}{\left(2k\right)!}=\frac{5!}{(2脳5)!}\left(Substituting\right) \\ =\frac{5!}{10!} \\ =\frac{5!}{10脳9脳8脳7脳6脳5!} \\ =\frac{1}{151200} \end{gathered}$$

The fifth term of the series $\underset{k=0}{\overset{鈭�}{鈭�}}\frac{k!}{\left(2k\right)!}$is $\frac{1}{151200}$.