问题
解答题
已知,一次函数y=
|
答案
令x=0,得y=
,y=0,得x=1 k+1
,1 k
∴S=
×1 2
×1 k+1
=1 k
(1 2
-1 k
),1 k+1
∴S1+S2+S3+…+S2012
=
(1-1 2
+1 2
-1 2
+1 3
-1 3
+…+1 4
-1 2012
)1 2013
=
(1-1 2
)1 2013
=
.1006 2013