题目内容
等差数列{an}，{bn}的前n项和分别为Sn和Tn，若
=
，则
=
．
| Sn |
| Tn |
| 2n |
| 3n+1 |
| lim |
| n→∞ |
| an |
| bn |
| 2 |
| 3 |
| 2 |
| 3 |
分析：利用等差数列的性质，结合等差数列的前n项和公式，可求
，进而可求
．
| an |
| bn |
| lim |
| n→∞ |
| an |
| bn |
解答：解：∵
=
，
∴
=
=
=
=
∴
=
=
=
故答案为：
| Sn |
| Tn |
| 2n |
| 3n+1 |
∴
| an |
| bn |
| a1+a2n-1 |
| b1+b2n-1 |
| S2n-1 |
| T2n-1 |
| 2(2n-1) |
| 3(2n-1)+1 |
| 2n-1 |
| 3n-1 |
∴
| lim |
| n→∞ |
| an |
| bn |
| lim |
| n→∞ |
| 2n-1 |
| 3n-1 |
| lim |
| n→∞ |
2-
| ||
3-
|
| 2 |
| 3 |
故答案为：
| 2 |
| 3 |
点评：本题考查等差数列的性质，等差数列的前n项和公式，考查数列的极限，正确求
是关键．
| an |
| bn |
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