爷爷是ee
幼苗
共回答了18个问题采纳率:83.3% 举报
根据已知的三个等式,总结规律得
1
n(n+1) =
1
n -
1
n+1 ,
(1)原式=
1
(x+1)(x+2) +
1
(x+2)(x+3) +
1
(x+3)(x+4)
=
1
x+1 -
1
x+2 +
1
x+2 -
1
x+3 +
1
x+3 -
1
x+4 =
1
x+1 -
1
x+4 =
3
(x+1)(x+4) ;
(2)由
a-1 + (ab-2) 2 =0 得:a-1=0且ab-2=0,
解得a=1且ab=2,
所以b=2,
则原式=
1
ab +
1
(a+1)(b+1) +…+
1
(a+2010)(b+2010) ,
=
1
1×2 +
1
2×3 +…+
1
2011×2012 ,
=1-
1
2 +
1
2 -
1
3 +
1
3 -
1
4 +…+
1
2010 -
1
2011 +
1
2011 -
1
2012 =1-
1
2012 =
2011
2012 .
故答案为:
1
n -
1
n+1 .
1年前
1