题目内容
已知数列{an},a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则
+
+
+…+
=( )
| 1 |
| S1 |
| 1 |
| S2 |
| 1 |
| S3 |
| 1 |
| Sn |
A.
| B.
| C.
| D.
|
∵点P(an,an+1)(n∈N*)在直线x-y+1=0上
∴an-an+1+1=0
∴数列{an}是以1为首项,以1为公差的等差数列.
∴an=n
∴sn=
∴
=
=2(
-
)
+
+
+…+
=2(1-
+
-
+…+
-
)=
故选C
∴an-an+1+1=0
∴数列{an}是以1为首项,以1为公差的等差数列.
∴an=n
∴sn=
| n(n+1) |
| 2 |
∴
| 1 |
| sn |
| 2 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| S1 |
| 1 |
| S2 |
| 1 |
| S3 |
| 1 |
| Sn |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
| 2n |
| n+1 |
故选C
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