有理数a,b满足|ab-2|+(1-b)2=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2008)(b+2008)的值.
解
|ab-2|+(1-b)²-0
∵|ab-2|≥0,(1-b)²≥0
∴ab-2=0,1-b=0
∴a=2,b=1
∴1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2008)(b+2008)
=1/1×2+1/2×3+1/3×4+…+1/2009×2010
=（1-1/2)+(1/2-1/3)+(1/3-1/4)+…+(1/2009-1/2010)
=1+(-1/2+1/2)+(-1/3+1/3)+…+(-1/2009+1/2009)-1/2010
=1+(-1/2010)
=2009/2010
1/1×2=1-1/2
1/2×3=1/2-1/3
.
.
1/2009×2010=1/2009-1/2009
你会发现中间的数,如-1/2与1/2,-1/3与1/3都可以相加变为0,只剩余
1和-1/2010
然后就可以计算出结果了!
