数列1,
|
由于an=
=1 1+2+…+n 1 n(n+1) 2
=
=2(2 n(n+1)
-1 n
)1 n+1
则S2009=2[(1-
)+(1 2
-1 2
)+…+(1 3
-1 2009
)]=2(1-1 2010
)=1 2010
=2×2009 2010 2009 1005
故答案为 A
数列1,
|
由于an=
=1 1+2+…+n 1 n(n+1) 2
=
=2(2 n(n+1)
-1 n
)1 n+1
则S2009=2[(1-
)+(1 2
-1 2
)+…+(1 3
-1 2009
)]=2(1-1 2010
)=1 2010
=2×2009 2010 2009 1005
故答案为 A