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数学问题啊

来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/05/01 09:56:31

解题思路: 证明: 假设当n=k时,A=1/(k+1)+1/(k+2)+…+1/(k+k)>13/24成立,则 当n=k+1时,左边=1/(k+2)+1/(k+3)+…+1/(k+1+k+1)=A+1/(k+1+k)+1/(k+1+k+1)-1/(k+1)=A+1/(2k+1)-1/(2k+2)=A+1/(2k+1)(2k+2)>A>13/24 即当n=k+1时,不等式仍成立 且当n=2时,不等式左边=1/3+1/4=7/12>13/24成立 ∴由归纳法可知不等式1/(n+1)+1/(n+2)+…+1/(n+n)>13/24成立
解题过程:
证明: 假设当n=k时,A=1/(k+1)+1/(k+2)+…+1/(k+k)>13/24成立,则 当n=k+1时,左边=1/(k+2)+1/(k+3)+…+1/(k+1+k+1)=A+1/(k+1+k)+1/(k+1+k+1)-1/(k+1)=A+1/(2k+1)-1/(2k+2)=A+1/(2k+1)(2k+2)>A>13/24 即当n=k+1时,不等式仍成立 且当n=2时,不等式左边=1/3+1/4=7/12>13/24成立 ∴由归纳法可知不等式1/(n+1)+1/(n+2)+…+1/(n+n)>13/24成立