问题 选择题
数列
1
2•5
1
5•8
1
8•11
1
(3n-1)(3n+2)
,…的前n项和Sn为(  )
A.
n
3n+2
B.
n
6n+4
C.
3n
6n+4
D.
n+1
n+2
答案

1
(3n-1)(3n+2)
1
3
(
1
3n-1
-
1
3n+2
)

Sn=

1
2•5
+
1
5•8
+…+
1
(3n-1)(3n+2)

=

1
3
(
1
2
-
1
5
+
1
5
-
1
8
+…+
1
3n-1
-
1
3n+2
)

=

1
3
(
1
2
-
1
3n+2
)=
n
6n+4

故选B

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