1. 练习题目

辅导初中学生数学的过程中，发现一道有意思的题目，分享如下。

1.1 题目描述

计算：
$$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\cdots +\sqrt{1+\frac{1}{{2010}^2}+\frac{1}{{2011}^2}}$$
本题是面向初二上学期的一道二次根式练习题，难度为“培优级”。

1.2 思路

这道题，应分析通项（即每一个根式项）。

2. 答题

2.1 分析通项

令通项为：
$$u_n=\sqrt{1+\frac{1}{n^2}+\frac{1}{{(n+1)}^2}}$$
则题目即计算
$$\sum _{n=1}^{2010}{u}_n$$
令
$$v_n=1+\frac{1}{n^2}+\frac{1}{{(n+1)}^2}$$
显然有：$$u_n=\sqrt{v_n}$$
我们努力化简 $v_n$，看看它能不能被开平方。
$$\begin{array}{rl}{v}_n& =1+\frac{1}{n^2}+\frac{1}{{\left(n+1\right)}^2}\\ & =1+\frac{{\left(n+1\right)}^2+n^2}{n^2\bullet {\left(n+1\right)}^2}\\ & =\frac{n^2\bullet {\left(n+1\right)}^2+{\left(n+1\right)}^2+n^2}{n^2\bullet {\left(n+1\right)}^2}\end{array}$$
令
$$w_n=n^2\bullet {\left(n+1\right)}^2+{\left(n+1\right)}^2+n^2$$
令
$$\begin{array}{rl}{w}_n& =n^2\bullet {\left(n+1\right)}^2+{\left(n+1\right)}^2+n^2\\ & =n^2\bullet \left(n^2+2n+1\right)+\left(n^2+2n+1\right)+n^2\\ & =\left(n^4+2n^3+n^2\right)+\left(n^2+2n+1\right)+n^2\\ & =n^2\bullet \left(n^2+2n+1\right)+\left(n^2+2n+1\right)+n^2\\ & =\left(n^4+2n^3+n^2\right)+\left(n^2+2n+1\right)+n^2\\ & =n^4+2n^3+{3n}^2+2n+1\\ & =\left(n^4+n^3+n^2\right)+\left(n^3+n^2+n\right)+\left(n^2+n+1\right)\\ & =n^2\bullet \left(n^2+n+1\right)+n\bullet \left(n^2+n+1\right)+\left(n^2+n+1\right)\\ & =\left(n^2+n+1\right)\left(n^2+n+1\right)\\ & ={\left(n^2+n+1\right)}^2\end{array}$$
则
$$v_n=\frac{w_n}{n^2\bullet {\left(n+1\right)}^2}=\frac{{\left(n^2+n+1\right)}^2}{n^2\bullet {\left(n+1\right)}^2}={\left[\frac{n^2+n+1}{n\bullet (n+1)}\right]}^2$$
则
$$u_n=\sqrt{v_n}=\frac{n^2+n+1}{n\bullet \left(n+1\right)}=\frac{n\bullet \left(n+1\right)+1}{n\bullet \left(n+1\right)}=1+\frac{1}{n\bullet \left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}$$

2.2 求通项的和

则
$$\begin{array}{rl}原式& =\sum _{n=1}^{2010}{u}_n\\ & =\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\\ & =2010+(\frac{1}{1}-\frac{1}{2011})=2011-\frac{1}{2011}\end{array}$$
上式的结果按照带分数来表示，即：
$$原式=2010\frac{2010}{2011}$$
【答毕】

鸣谢

初次尝试在 Markdown 文档中插入 KaTeX 数学公式，从以下链接中获得了帮助，特此感谢：

• KaTeX问题 —— csdn编辑时中打出等号对齐的样式
• KaTeX: Supported Functions
• word自带公式等号对齐（可任意符号处对齐）