则
| lim |
| n→∞ |
| an |
| bn |
| lim |
| n→∞ |
| a1 +(n−1)d1 |
| b1 +(n−1)d 2 |
| lim |
| n→∞ |
| ||||
|
| d1 |
| d2 |
∴
| lim |
| n→∞ |
| a1+a2+…+an |
| nb2n |
| lim |
| n→∞ |
na1 +
| ||
| n([b1 +(2n−1)d2] |
| ||
| 2d2 |
| 3 |
| 4 |
故答案为
| 3 |
| 4 |
| lim |
| n→∞ |
| an |
| bn |
| lim |
| n→∞ |
| a1+a2+…+an |
| nb2n |
| lim |
| n→∞ |
| an |
| bn |
| lim |
| n→∞ |
| a1 +(n−1)d1 |
| b1 +(n−1)d 2 |
| lim |
| n→∞ |
| ||||
|
| d1 |
| d2 |
| lim |
| n→∞ |
| a1+a2+…+an |
| nb2n |
| lim |
| n→∞ |
na1 +
| ||
| n([b1 +(2n−1)d2] |
| ||
| 2d2 |
| 3 |
| 4 |
| 3 |
| 4 |