题目内容
若数列{an}的通项公式为an=
,其前n项和为
,则n为( )
1 |
n2+3n+2 |
7 |
18 |
A、5 | B、6 | C、7 | D、8 |
考点:数列的求和
专题:等差数列与等比数列
分析:由已知得an=
=
=
-
,从而Sn=
-
+
-
+…+
-
=
-
,由此能求出结果.
1 |
n2+3n+2 |
1 |
(n+1)(n+2) |
1 |
n+1 |
1 |
n+2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n+1 |
1 |
n+2 |
1 |
2 |
1 |
n+2 |
解答:
解:∵an=
=
=
-
,
∴Sn=
-
+
-
+…+
-
=
-
,
∵前n项和为
,
∴
-
=
,
解得n=7.
故选:C.
1 |
n2+3n+2 |
1 |
(n+1)(n+2) |
1 |
n+1 |
1 |
n+2 |
∴Sn=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n+1 |
1 |
n+2 |
=
1 |
2 |
1 |
n+2 |
∵前n项和为
7 |
18 |
∴
1 |
2 |
1 |
n+2 |
7 |
18 |
解得n=7.
故选:C.
点评:本题考查满足条件的项数的求法,是中档题,解题时要认真审题,注意裂项求和法的合理运用.
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