题目内容
已知等差数列an满足:a3=7,a5+a7=26,令bn=
(n∈N*),则数列bn的前n项和Tn=______.
| 1 | ||
|
∵等差数列an满足:a3=7,a5+a7=26,
∴a3+a5+a7=33,
∴a5=11
∴d=
=2
∴an=2n+1,
∴bn=
=
∴4Tn=
+
+…+
=1-
+
-
+…+
-
=
∴Tn =
故答案为:
∴a3+a5+a7=33,
∴a5=11
∴d=
| 11-7 |
| 2 |
∴an=2n+1,
∴bn=
| 1 | ||
|
| 1 |
| 4n(n+1) |
∴4Tn=
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| n(n+1) |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
| n |
| n+1 |
∴Tn =
| n |
| 4(n+1) |
故答案为:
| n |
| 4(n+1) |
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