数学归纳法证明不等式问题(有图)最后两题.
数学归纳法证明不等式问题(有图)最后两题.
数学归纳法证明不等式问题(有图)
数学归纳法证明不等式问题(有图)最后两题.
10.
命题:n!≥2^(n-1)
n=1
1!=1
2^0=1
成立
n=k≥1时
假如有k!≥2^(k-1)
则n=k+1≥2时
(k+1)!=(k+1)*k!
≥2*k!
≥2*2^(k-1) (利用归纳假设)
≥2^k=2^[(k+1)-1]
所以对于任意n∈N*都有n!≥2^(n-1)
11.
n=1,1/2+1/3+1/4=26/24
n=2,1/3+1/4+...+1/7=153/140>26/24
所以猜测左边的数随着n增大而增加,所以a/2425/24
n=1时已验证成立
假设n=k时成立
则
1/(k+1)+1/(k+2)+...+1/(3k+1)>25/24
n=k+1时
1/(k+1+1)+1/(k+1+2)+...+1/(3(k+1)+1)
=1/(k+2)+1/(k+3)+...+1/(3k+4)
=[1/(k+1)+1/(k+2)+...+1/(3k+1)]+1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
=[1/(k+1)+1/(k+2)+...+1/(3k+1)]+1/(3k+2)+1/(3k+4)-2/(3k+3)
=[1/(k+1)+1/(k+2)+...+1/(3k+1)]+[1/(3k+2)-1/(3k+3)]-[1/(3k+3)-1/(3k+4)]
=[1/(k+1)+1/(k+2)+...+1/(3k+1)]+1/[(3k+2)(3k+3)]-1/[(3k+3)(3k+4)]
=[1/(k+1)+1/(k+2)+...+1/(3k+1)]+2/[(3k+2)(3k+3)(3k+4)]
>[1/(k+1)+1/(k+2)+...+1/(3k+1)]
>25/24
所以对于所有n∈N*都有1/(n+1)+1/(n+2)+...+1/(3n+1)>25/24
第一个做出来了。。等等我给你写。。。第二个我再看看。好。这类题目怎么想比较好。前两题我也是看了同桌的恍然大悟。老师说要擅长用放缩法。可我放缩法不怎么会啊