Question

已知|ab-2|与|a-1|互为相反数,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……

Answer 1

因为|ab-2|和|a-1|互为相反数,所以二者都为0,推出a=1,b=2

所以原式=1/(1*2)+1/(2*3)+.+1/(2005*2006)

=1-1/2+1/2-1/3+.+1/2005-1/2006 (约去中间项)

=1-1/2006

=2005/2006

这一类题目都是这么用拆项法解.