计算(1-
|
设(1-
-1 2
-…-1 3
)=x;(1 2003
+1 2
+…+1 3
)=y1 2003
则原式=x(y+
)-(x-1 2004
)y1 2004
=xy+
-xy+x 2004 y 2004
=x+y 2004
又∵x+y=(1-
-1 2
-…-1 3
)+(1 2003
+1 2
+…+1 3
)=11 2003
∴原式=1 2004
故选A.
计算(1-
|
设(1-
-1 2
-…-1 3
)=x;(1 2003
+1 2
+…+1 3
)=y1 2003
则原式=x(y+
)-(x-1 2004
)y1 2004
=xy+
-xy+x 2004 y 2004
=x+y 2004
又∵x+y=(1-
-1 2
-…-1 3
)+(1 2003
+1 2
+…+1 3
)=11 2003
∴原式=1 2004
故选A.