# 109中壢高中代理

\begin{align} & {{k}_{2}}=1\ ,\ 1853\le {{k}_{1}}\le 2020 \\ & : \\ & : \\ & {{k}_{2}}=167\ ,\ 2019\le {{k}_{1}}\le 2186 \\ \end{align}

TOP

TOP

\begin{align} & {{k}_{2}}=1852\quad ,\quad 1853\le {{k}_{1}}\le 2019 \\ & {{k}_{2}}=1853\quad ,\quad 1854\le {{k}_{1}}\le 2019 \\ & : \\ & : \\ & {{k}_{2}}=2017\quad ,\quad 2018\le {{k}_{1}}\le 2019 \\ & {{k}_{2}}=2018\quad ,\quad {{k}_{1}}=2019 \\ \end{align}

TOP

TOP

## 回復 14# anyway13 的帖子

\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}

TOP

﻿