2014.2.1.用数学归纳法证明：第一题只要从k+1这一步开始证明就可以了!

1、n=1时,1/2^2=1/4,1/2-1/(1+2)=1/61/2-1/(k+2),
3、1/2^2+1/3^2+...+1/(k+1)^2+1/(k+2)^2
>1/2-1/(k+2)+1/(k+2)^2
=1/2-(k+1)/(k+2)^2
=1/2-1/(k+3)+[1/(k+3)-(k+1)/(k+2)^2]
=1/2-1/(k+3)+[(k+2)^2-(k+1)(k+3)]/(k+3)(k+2)^2
=1/2-1/(k+3)+1/(k+3)(k+2)^2
>1/2-1/(k+3),
即n=k+1时,原命题也成立,命题得证.