题目内容
数列{an}中,an=
,Sn为{an}的前n项和,则S1+S2+…+S10的值为( )
| 1 |
| n(n+1)(n+2) |
A.
| B.
| C.
| D.
|
∵an=
[(
-
)-(
-
)],
∴Sn=
[1-
-(
-
)]=
+
(
-
).
∴S1+S2+…+S10=
+
[(
-
)+(
-
)+(
-
)+…+(
-
)]
=
+
(
-
)
=
.
故选A.
| 1 |
| 2 |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| n+1 |
| 1 |
| n+2 |
∴Sn=
| 1 |
| 2 |
| 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.
练习册系列答案
相关题目