Question

用数学归纳法证明

证明1/(n+1)+1/(n+2)+....+1/(3n+1)≥1 （n∈N）

Answer 1

证明：
当n=1时，1/2 + 1/3 +1/4=13/12＞1，结论成立。
令An=1/(n+1)+1/(n+2)+....+1/(3n+1)
假设当n=k时结论成立，即
Ak=1/(k+1)+1/(k+2)+…+1/(3k+1)>1
我们来证明n=k+1时，结论也成立
因为
A(k+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)
=Ak +1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
下面我们来证明1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)＞0 ①
①式可化左端可化为
1/(3k+3-1)+1/(3k+3)+1/(3k+3+1)-3/(3k+3)
=1/(3k+3-1)+1/(3k+3+1)-2/(3k+3) ②
令a=3k+3
若1/（a-1) +1/(a+1)＞2/a (其中a＞1) 成立
则②>0
1/（a-1) +1/(a+1)=2a/(a²-1)＞2a/a²=2/a
这样1/（a-1) +1/(a+1)＞2/a成立，从而②式大于0，即①式成立，从而
A(k+1)＞Ak＞1

Answer 2

n=1时，左边=1/2+1/3+1/4=6/12+4/12+3/12=13/12>1
设n=k时成立，即：1/(k+1)+1/(k+2)+....+1/(3k+1)≥1 ，则
n=k+1时，原式左边为：1/(k+2)+1/(k+3)+....+1/(3k+1）+1/(3k+2）+1/(3k+3）+1/(3k+4）
=1/(k+1)+1/(k+2)+....+1/(3k+1）-1/(k+1)+1/(3k+2）+1/(3k+3）+1/(3k+4）
=1/(k+1)+1/(k+2)+....+1/(3k+1）+1/(3k+2）+1/(3k+3）+1/(3k+4）-3/(3k+3)
=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））
显然后面部分是大于0的，故原式得证