题目内容
已知数列{an}中,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 |
∵点P(an,an+1)(n∈N*)在直线x-y+1=0上,
∴an+1-an=1,
∴数列{an}是等差数列,
∵a1=1,
∴sn=
,
∴
=
,
∴
+
+
+…+
=2(1-
+
-…-
)=
,
故答案为
.
∴an+1-an=1,
∴数列{an}是等差数列,
∵a1=1,
∴sn=
| n2+n |
| 2 |
∴
| 1 |
| sn |
| 2 |
| n(n+1) |
∴
| 1 |
| S1 |
| 1 |
| S2 |
| 1 |
| S3 |
| 1 |
| Sn |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| n+1 |
| 2n |
| n+1 |
故答案为
| 2n |
| n+1 |
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