Question

用数学归纳法证明：1/n+1/(1+n)+1/(n+2) +......1/n^2>1(n∈N且n>1)

Answer 1

证明:
(1)当n=2,
1/2+1/3+1/4=13/12>1成立
(2)假设当n=k时,即
1/k+1/(k+1)+...+1/k^2>1
所以当n=k+1时,有:
1/(k+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)
=1/k+1/(k+1)+...+1/k^2+[1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)-1/k]
>1+[1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1)-1/k]
>1+[(2k+1)/(k^2+2k+1)-1/k]
=1+[(2k²+k-k²-2k-1)/(k²+2k+1)k]
=1+[(k²-k-1)/(k²+2k+1)k]
因为:
k²-k-1＞0（当k＞2时）
(k²-k-1)/(k²+2k+1)k>0
所以:
1/(k+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)
>1+0
=1
所以当n=k+1原式也成立
综上,有:
1/n+1/(n+1)+1/(n+2)+…+1/n^2>1(n>1且n是整数)

Answer 2

n=1时，左边=1*1=1

右边=1/6*1*2*3=1

左边=右边，等式成立！
假设n=k时成立
(k>1)即:
1*k+2(k-1)+3(k-2)+…+(k-1)*2+k*1=(1/6)k(k+1)(k+2)
当n=k+1时；
左边
=1*(k+1)+2(k+1-1)+3(k+1-2)+…+(k+1-1)*2+(k+1)*1
=1*k+1*1+2(k-1)+2*1+…+k*1+k+(k+1)
=[1*k+2(k-1)+…+(k-1)*2+k*1]+1+2+3+…+k+(k+1)
=(1/6)k(k+1)(k+2)+1+2+3+…+k+(k+1)
=(1/6)k(k+1)(k+2)+1/2*(k+1)*(k+2)
=(1/6)(k+1)(k+2)(k+3)
=(1/6)(k+1)[(k+1)+1][(k+1)+2]
=右边
原式也成立！
综上可知，原式为真！