| 1+2+…+n |
| n |
| n(n+1) |
| 2n |
| n+1 |
| 2 |
| 1 |
| anan+1 |
| 4 |
| (n+1)(n+2) |
| 1 |
| n+1 |
| 1 |
| n+2 |
∴数列{bn}的前n项和=4[(
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n+1 |
| 1 |
| n+2 |
=4(
| 1 |
| 2 |
| 1 |
| n+2 |
| 2n |
| n+2 |
故答案为
| 2n |
| n+2 |
| 1+2+…+n |
| n |
| 1 |
| anan+1 |
| 1+2+…+n |
| n |
| n(n+1) |
| 2n |
| n+1 |
| 2 |
| 1 |
| anan+1 |
| 4 |
| (n+1)(n+2) |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2 |
| 1 |
| n+2 |
| 2n |
| n+2 |
| 2n |
| n+2 |