代入an+1-3an=3n中，得3n+1bn+1-3n+1bn=3n，
即得bn+1?bn＝
| 1 |
| 3 |
所以数列{bn}是等差数列．（6分）
（2）解：因为数列{bn}是首项为b1=3-1a1=1，公差为
| 1 |
| 3 |
则bn＝1+
| 1 |
| 3 |
| n+2 |
| 3 |
从而有
| an |
| n+2 |
故Sn＝
| a1 |
| 3 |
| a2 |
| 4 |
| a3 |
| 5 |
| an |
| n+2 |
| 1?3n |
| 1?3 |
| 3n?1 |
| 2 |
则
| Sn |
| S2n |
| 3n?1 |
| 32n?1 |
| 1 |
| 3n+1 |
由
| 1 |
| 128 |
| Sn |
| S2n |
| 1 |
| 4 |
| 1 |
| 128 |
| 1 |
| 3n+1 |
| 1 |
| 4 |
即3＜3n＜127，得1＜n≤4．
故满足不等式
| 1 |
| 128 |
| Sn |
| S2n |
| 1 |
| 4 |
