题目详情 - Q20260201111052929
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