回復 14# anyway13 的帖子
第19題
\(\begin{align}
& \frac{3}{2!}-\frac{4}{3!}+\frac{5}{4!}-\frac{6}{5!}+\frac{7}{6!}-\cdots +{{\left( -1 \right)}^{n+1}}\times \frac{n+2}{\left( n+1 \right)!} \\
& =\left[ \frac{2}{2!}-\frac{3}{3!}+\frac{4}{4!}-\frac{5}{5!}+\frac{6}{6!}-\cdots +{{\left( -1 \right)}^{n+1}}\times \frac{n+1}{\left( n+1 \right)!} \right]+\left[ \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\cdots +{{\left( -1 \right)}^{n+1}}\times \frac{1}{\left( n+1 \right)!} \right] \\
& =\left[ 1-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{5!}-\cdots +{{\left( -1 \right)}^{n+1}}\times \frac{1}{n!} \right]+\left[ \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\cdots +{{\left( -1 \right)}^{n+1}}\times \frac{1}{\left( n+1 \right)!} \right] \\
& =1 \\
\end{align}\)