| an |
| 2n |
| an−1 |
| 2n−1 |
∴
| an |
| 2n |
| an−1 |
| 2n−1 |
又bn＝
| an |
| 2n |
∴数列{bn}是首项为1，公差为2的等差数列．…（6分）
（2）由（1）知bn=2n-1，∴
| 1 |
| bnbn+1 |
| 1 |
| (2n−1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n−1 |
| 1 |
| 2n+1 |
∴Tn＝
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2n−1 |
| 1 |
| 2n+1 |
| 1 |
| 2 |
| 1 |
| 2n+1 |
| n |
| 2n+1 |
| an |
| 2n |
| 1 |
| b1b2 |
| 1 |
| b2b3 |
| 1 |
| bnbn+1 |
| an |
| 2n |
| an−1 |
| 2n−1 |
| an |
| 2n |
| an−1 |
| 2n−1 |
| an |
| 2n |
| 1 |
| bnbn+1 |
| 1 |
| (2n−1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n−1 |
| 1 |
| 2n+1 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2n−1 |
| 1 |
| 2n+1 |
| 1 |
| 2 |
| 1 |
| 2n+1 |
| n |
| 2n+1 |