简介

表述 1

$a_n=\phantom{0}$$\phantom{0}{\displaystyle \frac{b}{(kn+p)(kn+q)}}$。若 $a_n$ 满足

• $(kn+\phantom{0}$$\phantom{0}p)(kn+\phantom{0}$$\phantom{0}q)=\phantom{0}$$\phantom{0}0$ 的两根 $n_1-\phantom{0}$$\phantom{0}{n}_2=\phantom{0}$$\phantom{0}1$；
• 大根 $n_1$ 所在的位置是 $(kn+\phantom{0}$$\phantom{0}p)$，即 $k{n}_1+\phantom{0}$$\phantom{0}p=\phantom{0}$$\phantom{0}0$。

则 $\{a_n\}$ 的前 $n$ 项和为

$$S_n={\displaystyle \frac{bn}{(k+p)(kn+q)}}$$

若 $n_1-\phantom{0}$$\phantom{0}{n}_2\ne \phantom{0}$$\phantom{0}1$ 则不可使用 N1 互换。

表述 2

$a_n$ 为等差数列，则 ${\displaystyle \frac{1}{a_n{a}_{n+1}}}$ 的前 $n$ 项和为

$$S_n={\displaystyle \frac{n}{a_1{a}_{n+1}}}$$

容易证明同 表述 1 等价。

原理

$$\begin{array}{rl}{\displaystyle \frac{1}{a_n{a}_{n+1}}}& ={\displaystyle \frac{1}{d}}\cdot {\displaystyle \frac{a_{n+1}-a_n}{a_n{a}_{n+1}}}={\displaystyle \frac{1}{d}}({\displaystyle \frac{1}{a_n}}-{\displaystyle \frac{1}{a_{n+1}}})\\ S_n& ={\displaystyle \frac{1}{d}}({\displaystyle \frac{1}{a_1}}-{\displaystyle \frac{1}{a_2}}+{\displaystyle \frac{1}{a_2}}-{\displaystyle \frac{1}{a_3}}+\cdots +{\displaystyle \frac{1}{a_n}}-{\displaystyle \frac{1}{a_{n+1}}})\\ & ={\displaystyle \frac{1}{d}}({\displaystyle \frac{1}{a_1}}-{\displaystyle \frac{1}{a_{n+1}}})\\ & ={\displaystyle \frac{1}{d}}\cdot {\displaystyle \frac{a_{n+1}-a_1}{a_1{a}_{n+1}}}\\ & ={\displaystyle \frac{n}{a_1{a}_{n+1}}}\end{array}$$

例 1

已知 $a_n=\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{(2n+1)(2n+3)}}$，则 $\{a_n\}$ 的前 $n$ 项和 $S_n=\phantom{0}$$\phantom{0}\underset{―}{whatthefuck}$。

常规解法

1.裂项

$$a_n={\displaystyle \frac{1}{(2n+1)(2n+3)}}={\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{2n+1}}-{\displaystyle \frac{1}{2n+3}})$$

2.求和

$$\begin{array}{rl}{S}_n& =a_1+a_2+a_3+\cdots +a_n\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{7}}+\cdots +{\displaystyle \frac{1}{2n+1}}-{\displaystyle \frac{1}{2n+3}})\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{2n+3}})\\ & ={\displaystyle \frac{n}{3(2n+3)}}\end{array}$$

解

令 $(2n+\phantom{0}$$\phantom{0}1)(2n+\phantom{0}$$\phantom{0}3)=\phantom{0}$$\phantom{0}0,\phantom{0}$$\phantom{0}{n}_1-\phantom{0}$$\phantom{0}{n}_2=\phantom{0}$$\phantom{0}-\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}-\phantom{0}$$\phantom{0}(-{\displaystyle \frac{3}{2}})=\phantom{0}$$\phantom{0}1$，可用 N1 互换。

$$S_n={\displaystyle \frac{n}{(2+1)(2n+3)}}={\displaystyle \frac{n}{3(2n+3)}}$$

例 2

已知 $a_n=\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{4n^2-1}}$，则 $\{a_n\}$ 的前 $n$ 项和 $S_n=\phantom{0}$$\phantom{0}\underset{―}{whatthefuck}$。

解

$a_n=\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{(2n-1)(2n+1)}}$。

令 $(2n-\phantom{0}$$\phantom{0}1)(2n+\phantom{0}$$\phantom{0}1)=\phantom{0}$$\phantom{0}0,\phantom{0}$$\phantom{0}{n}_1-\phantom{0}$$\phantom{0}{n}_2=\phantom{0}$$\phantom{0}{\displaystyle \frac{1}{2}}-\phantom{0}$$\phantom{0}(-{\displaystyle \frac{1}{2}})=\phantom{0}$$\phantom{0}1$，可用 N1 互换。

$$S_n={\displaystyle \frac{n}{(2-1)(2n+1)}}={\displaystyle \frac{n}{2n+1}}$$

例 3

已知 $a_n=\phantom{0}$$\phantom{0}{\displaystyle \frac{2}{n(n+1)}}$，则 $\{a_n\}$ 的前 $n$ 项和 $S_n=\phantom{0}$$\phantom{0}\underset{―}{whatthefuck}$。

解

令 $n(n+\phantom{0}$$\phantom{0}1)=\phantom{0}$$\phantom{0}0,\phantom{0}$$\phantom{0}{n}_1-\phantom{0}$$\phantom{0}{n}_2=\phantom{0}$$\phantom{0}0-\phantom{0}$$\phantom{0}(-1)=\phantom{0}$$\phantom{0}1$，可用 N1 互换。

$$S_n={\displaystyle \frac{2n}{n+1}}$$