问题
解答题
已知
观察:
|
答案
∵
+(ab-2)2=0,a-1
∴
,解得a-1=0 ab-2=0
,a=1 b=2
∴原式=
+1 2
+1 2×3
+…+1 3×4 1 2011×2012
=
+1 2
-1 2
+1 3
-1 3
+…+1 4
-1 2011 1 2012
=1-1 2012
=
.2011 2012
已知
观察:
|
∵
+(ab-2)2=0,a-1
∴
,解得a-1=0 ab-2=0
,a=1 b=2
∴原式=
+1 2
+1 2×3
+…+1 3×4 1 2011×2012
=
+1 2
-1 2
+1 3
-1 3
+…+1 4
-1 2011 1 2012
=1-1 2012
=
.2011 2012