7. Find the value of $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\cdots+\frac{22}{20!+21!+22!}$.
(A) $1-\frac{1}{24!}$
(B) $\frac{1}{2}-\frac{1}{23!}$
(C) $\frac{1}{2}-\frac{1}{22!}$
(D) $1-\frac{1}{22!}$
(E) $\frac{1}{2}-\frac{1}{24!}$
7. Answer: (C)
First note that $(n-2)!+(n-1)!+n!=(n-2)![1+(n-1)+n(n-1)]=n^{2}(n-2)!$.
Therefore the given series can be written as
$$
\begin{aligned}
\sum_{n=3}^{22} \frac{n}{n^{2}(n-2)!} & =\sum_{n=3}^{22} \frac{1}{n(n-2)!}=\sum_{x=3}^{22} \frac{n-1}{n(n-1)(n-2)!} \\
& =\sum_{n=3}^{22} \frac{n-1}{n!}=\sum_{n=3}^{22}\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right) .
\end{aligned}
$$

Summing the telescoping series, we obtain $\frac{1}{2!}-\frac{1}{22!}$.