题目内容
等差数列{an}中，已知an=3n-1，若数列{
}的前n项和为
，则n的值为（ ）
| 1 |
| anan+1 |
| 4 |
| 25 |
分析：由an=3n-1，知
=
=
(
-
)，再由数列{
}的前n项和为
，利用裂项求和法建立方程即能求出n的值．
| 1 |
| anan+1 |
| 1 |
| (3n-1)(3n+2) |
| 1 |
| 3 |
| 1 |
| 3n-1 |
| 1 |
| 3n+2 |
| 1 |
| anan+1 |
| 4 |
| 25 |
解答：解：∵an=3n-1，
∴
=
=
(
-
)，
∵数列{
}的前n项和为
，
∴Sn=
(
-
)+
(
-
)+
(
-
)+…+
(
-
)
=
(
-
)=
，
解得n=16．
故选D．
∴
| 1 |
| anan+1 |
| 1 |
| (3n-1)(3n+2) |
| 1 |
| 3 |
| 1 |
| 3n-1 |
| 1 |
| 3n+2 |
∵数列{
| 1 |
| anan+1 |
| 4 |
| 25 |
∴Sn=
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 8 |
| 1 |
| 3 |
| 1 |
| 8 |
| 1 |
| 11 |
| 1 |
| 3 |
| 1 |
| 3n-1 |
| 1 |
| 3n+2 |
=
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3n+2 |
| 4 |
| 25 |
解得n=16．
故选D．
点评：本题考查数列求和的应用，是中档题．解题时要认真审题，仔细解答，注意裂项求和法的合理运用．
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