数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10 n>=2

题目

数学归纳法证明1/n+1+1/n+2+1/n+3+...+1/3n>9/10 n>=2
1）当n=2时,左=1/3 +1/4+1/5+1/6=57/60>54/60=9/10,成立．
（2）假设n=k时,有1/(k+1) +1/(k+2) +...+1/3k >9/10
那么　1/(k+2)+1/(k+3) +...+1/3(k+1)
=[1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1) +1/(3k+2)+1/(3k+3) -1/(k+1)
>9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1)
=9/10
即n=k+1时命题也成立,
从而　原不等式对n∈N,且n>1成立．
第二步中为什么是
>9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1)
不应该是
>9/10 +1/(3k+1) +1/(3k+2)+1/(3k+3) -1/(k+1)的么

答案