求证:1n+1
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求证:
1 |
n+1 |
证明:(1)当n=2时,左边=
1
3+
1
4+
1
5+
1
6=
57
60>
50
60=
5
6,不等式成立;
(2)假设n=k(k≥2,k∈N*)时命题成立,即
1
k+1+
1
k+2+…+
1
3k>
5
6成立.
则当n=k+1时,左边=
1
(k+1)+1+
1
(k+1)+2+…+
1
3k+
1
3k+1+
1
3k+2+
1
3(k+1)
=
1
k+1+
1
k+2+…+
1
3k+(
1
3k+1+
1
3k+2+
1
3k+3−
1
k+1)
>
5
6+(3×
1
3k+3−
1
k+1)=
5
6.
所以当n=k+1时不等式也成立.
综上由(1)(2)可知:原不等式对任意n≥2(n∈N*)都成立.
1
3+
1
4+
1
5+
1
6=
57
60>
50
60=
5
6,不等式成立;
(2)假设n=k(k≥2,k∈N*)时命题成立,即
1
k+1+
1
k+2+…+
1
3k>
5
6成立.
则当n=k+1时,左边=
1
(k+1)+1+
1
(k+1)+2+…+
1
3k+
1
3k+1+
1
3k+2+
1
3(k+1)
=
1
k+1+
1
k+2+…+
1
3k+(
1
3k+1+
1
3k+2+
1
3k+3−
1
k+1)
>
5
6+(3×
1
3k+3−
1
k+1)=
5
6.
所以当n=k+1时不等式也成立.
综上由(1)(2)可知:原不等式对任意n≥2(n∈N*)都成立.
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