题目详情 - Q20260201111052929

S09_001_002 下第 32 / 107 题
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Q20260201111052929

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题干

设 $a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}\ (n\in N^*)$,那么 $a_{n+1}-a_n$ 等于( )

选项

A
$\frac{1}{2n+1}$
B
$\frac{1}{2n+2}$
C
$\frac{1}{2n+1}+\frac{1}{2n+2}$
D
$\frac{1}{2n+1}-\frac{1}{2n+2}$

正确答案

D

解析

$\because a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}$ $\therefore a_{n+1}=\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n+1}+\frac{1}{2n+2}$, $\therefore a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$ $=\frac{1}{2n+1}-\frac{1}{2n+2}$.
审核状态: 合格
S09_001_002