初中数学选讲：二次根式练习题（20240706-01）
- [1. 练习题目](#1. 练习题目)
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- [1.1 题目描述](#1.1 题目描述)
- [1.2 思路](#1.2 思路)
- [2. 答题](#2. 答题)
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- [2.1 分析通项](#2.1 分析通项)
- [2.2 求通项的和](#2.2 求通项的和)
- 鸣谢
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：
1 + 1 1 2 + 1 2 2 + 1 + 1 2 2 + 1 3 2 + 1 + 1 3 2 + 1 4 2 + ⋯ + 1 + 1 2010 2 + 1 2011 2 \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}} 1+121+221 +1+221+321 +1+321+421 +⋯+1+201021+201121
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 = 1 + 1 n 2 + 1 ( n + 1 ) 2 u_n=\sqrt{1+\frac{1}{n^2}+\frac{1}{{(n+1)}^2}} un=1+n21+(n+1)21
则题目即计算
∑ n = 1 2010 u n \sum_{n=1}^{2010}u_n n=1∑2010un
令
v n = 1 + 1 n 2 + 1 ( n + 1 ) 2 v_n=1+\frac{1}{n^2}+\frac{1}{{(n+1)}^2} vn=1+n21+(n+1)21
显然有： u n = v n u_n=\sqrt{v_n} un=vn
我们努力化简 v n v_n vn，看看它能不能被开平方。
v n = 1 + 1 n 2 + 1 ( n + 1 ) 2 = 1 + ( n + 1 ) 2 + n 2 n 2 ∙ ( n + 1 ) 2 = n 2 ∙ ( n + 1 ) 2 + ( n + 1 ) 2 + n 2 n 2 ∙ ( n + 1 ) 2 \begin{aligned} 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{aligned} vn=1+n21+(n+1)21=1+n2∙(n+1)2(n+1)2+n2=n2∙(n+1)2n2∙(n+1)2+(n+1)2+n2
令
w n = n 2 ∙ ( n + 1 ) 2 + ( n + 1 ) 2 + n 2 w_n=n^2\bullet\left(n+1\right)^2+\left(n+1\right)^2+n^2 wn=n2∙(n+1)2+(n+1)2+n2
则有
w n = n 2 ∙ ( n + 1 ) 2 + ( n + 1 ) 2 + n 2 = n 2 ∙ ( n 2 + 2 n + 1 ) + ( n 2 + 2 n + 1 ) + n 2 = ( n 4 + 2 n 3 + n 2 ) + ( n 2 + 2 n + 1 ) + n 2 = n 2 ∙ ( n 2 + 2 n + 1 ) + ( n 2 + 2 n + 1 ) + n 2 = ( n 4 + 2 n 3 + n 2 ) + ( n 2 + 2 n + 1 ) + n 2 = n 4 + 2 n 3 + 3 n 2 + 2 n + 1 = ( n 4 + n 3 + n 2 ) + ( n 3 + n 2 + n ) + ( n 2 + n + 1 ) = n 2 ∙ ( n 2 + n + 1 ) + n ∙ ( n 2 + n + 1 ) + ( n 2 + n + 1 ) = ( n 2 + n + 1 ) ( n 2 + n + 1 ) = ( n 2 + n + 1 ) 2 \begin{aligned} 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{aligned} wn=n2∙(n+1)2+(n+1)2+n2=n2∙(n2+2n+1)+(n2+2n+1)+n2=(n4+2n3+n2)+(n2+2n+1)+n2=n2∙(n2+2n+1)+(n2+2n+1)+n2=(n4+2n3+n2)+(n2+2n+1)+n2=n4+2n3+3n2+2n+1=(n4+n3+n2)+(n3+n2+n)+(n2+n+1)=n2∙(n2+n+1)+n∙(n2+n+1)+(n2+n+1)=(n2+n+1)(n2+n+1)=(n2+n+1)2
则
v n = w n n 2 ∙ ( n + 1 ) 2 = ( n 2 + n + 1 ) 2 n 2 ∙ ( n + 1 ) 2 = [ n 2 + n + 1 n ∙ ( n + 1 ) ] 2 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 vn=n2∙(n+1)2wn=n2∙(n+1)2(n2+n+1)2=[n∙(n+1)n2+n+1]2
则
u n = v n = n 2 + n + 1 n ∙ ( n + 1 ) = n ∙ ( n + 1 ) + 1 n ∙ ( n + 1 ) = 1 + 1 n ∙ ( n + 1 ) = 1 + 1 n − 1 n + 1 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} un=vn =n∙(n+1)n2+n+1=n∙(n+1)n∙(n+1)+1=1+n∙(n+1)1=1+n1−n+11
2.2 求通项的和
则
原式 = ∑ n = 1 2010 u n = ( 1 + 1 1 − 1 2 ) + ( 1 + 1 2 − 1 3 ) + ( 1 + 1 3 − 1 4 ) + ⋯ + ( 1 + 1 2010 − 1 2011 ) = 2010 + ( 1 1 − 1 2011 ) = 2011 − 1 2011 \begin{aligned} 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{aligned} 原式=n=1∑2010un=(1+11−21)+(1+21−31)+(1+31−41)+⋯+(1+20101−20111)=2010+(11−20111)=2011−20111
上式的结果按照带分数来表示，即：
原式 = 2010 2010 2011 原式=2010\frac{2010}{2011} 原式=201020112010
【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：