y= (n^2+n)x^2-(2n+1)x+1
与x轴交,=> y=0
(n^2+n)x^2-(2n+1)x+1=0
anbn = 1/(n^2+n) >0
|anbn| = anbn = 1/n-1/(n+1)
|a1b1|+|a2b2|+...+|anbn| = 1-1/(n+1)
|a1b1|+|a2b2|+...+|a2012b2012| = 2012/2013你是怎么去掉绝对值符号的anbn = 1/(n^2+n) >0|anbn|=anbn