题目内容

数列{an}中,an=
1
n(n+1)(n+2)
,Sn为{an}的前n项和,则S1+S2+…+S10的值为(  )
A.
55
24
B.
1
24
C.
55
2
D.
65
24
an=
1
2
[(
1
n
-
1
n+1
)-(
1
n+1
-
1
n+2
)]

Sn=
1
2
[1-
1
n+1
-(
1
2
-
1
n+2
)]
=
1
4
+
1
2
(
1
n+2
-
1
n+1
)

∴S1+S2+…+S10=
10
4
+
1
2
[(
1
3
-
1
2
)+(
1
4
-
1
3
)+(
1
5
-
1
4
)+…+(
1
12
-
1
11
)]

=
5
2
+
1
2
(
1
12
-
1
2
)

=
55
24

故选A.
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