题目内容
已知数列{an}的通项公式an=
，求它的前n项和．
| (2n)2 |
| (2n-1)(2n+1) |
∵an=
+
，
∴Sn=(1+
)+(
+
)+…+(
+
)+(
+
)
=1+(
+
)+(
+
)+…+(
+
)+
=n+
=
．
| n |
| 2n-1 |
| n |
| 2n+1 |
∴Sn=(1+
| 1 |
| 3 |
| 2 |
| 3 |
| 2 |
| 5 |
| n-1 |
| 2n-3 |
| n-1 |
| 2n-1 |
| n |
| 2n-1 |
| n |
| 2n+1 |
=1+(
| 1 |
| 3 |
| 2 |
| 3 |
| 2 |
| 5 |
| 3 |
| 5 |
| n-1 |
| 2n-1 |
| n |
| 2n-1 |
| n |
| 2n+1 |
| n |
| 2n+1 |
=
| 2n(n+1) |
| 2n+1 |
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已知数列{an}的通项为an=2n-1，Sn为数列{an}的前n项和，令bn=
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| 1 |
| Sn+n |
A、[
| ||||
B、(
| ||||
C、[
| ||||
D、[
|