由a1=3，a3=9得2（log22+d）=log22+log28，即d=1．
所以log2（an-1）=1+（n-1）×1=n，即an=2n+1．
（II）证明：因为
| 1 |
| an+1−an |
| 1 |
| 2n+1−2n |
| 1 |
| 2n |
所以
| 1 |
| a2−a1 |
| 1 |
| a3−a2 |
| 1 |
| an+1−an |
| 1 |
| 21 |
| 1 |
| 22 |
| 1 |
| 23 |
| 1 |
| 2n |
| ||||||
1−
|
| 1 |
| 2n |
即得证．
| 1 |
| a2−a1 |
| 1 |
| a3−a2 |
| 1 |
| an+1−an |
| 1 |
| an+1−an |
| 1 |
| 2n+1−2n |
| 1 |
| 2n |
| 1 |
| a2−a1 |
| 1 |
| a3−a2 |
| 1 |
| an+1−an |
| 1 |
| 21 |
| 1 |
| 22 |
| 1 |
| 23 |
| 1 |
| 2n |
| ||||||
1−
|
| 1 |
| 2n |