题目内容
数列{an}中,an=
,bn=
的前n项和为______.
1+2+3+…+n |
n |
1 |
anan+1 |
设数列bn的前n项和为Sn
由题意可得an=
=
=
∴an+1=
∴bn=
=
=
=4(
-
)
∴Sn=b1+b2+…+bn-1+bn
=4(
-
+
-
…+
-
+
-
)
=4(
-
)
=
∴bn=
的前n项和为
.
由题意可得an=
1+2+3+…+n |
n |
| ||
n |
n+1 |
2 |
∴an+1=
n+2 |
2 |
∴bn=
1 |
anan+1 |
1 | ||||
|
4 |
(n+1)(n+2) |
1 |
n+1 |
1 |
n+2 |
∴Sn=b1+b2+…+bn-1+bn
=4(
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n |
1 |
n+1 |
1 |
n+1 |
1 |
n+2 |
=4(
1 |
2 |
1 |
n+2 |
=
2n |
n+2 |
∴bn=
1 |
anan+1 |
2n |
n+2 |
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