| a2 |
| an−1 |
∴bn=
| 1 |
| an−a |
| 1 | ||
a−
|
| an−1 |
| a(an−1−a) |
∴bn-bn-1=
| an−1 |
| a(an−1−a) |
| 1 |
| an−1−a |
| 1 |
| a |
∴数列{bn}是公差为
| 1 |
| a |
（2）∵b1=
| 1 |
| a1−a |
| 1 |
| a |
故由（1）得：bn=
| 1 |
| a |
| 1 |
| a |
| n |
| a |
即：
| 1 |
| an−a |
| n |
| a |
得：an=a（1+
| 1 |
| n |
| a2 |
| an−1 |
| 1 |
| an−a |
| a2 |
| an−1 |
| 1 |
| an−a |
| 1 | ||
a−
|
| an−1 |
| a(an−1−a) |
| an−1 |
| a(an−1−a) |
| 1 |
| an−1−a |
| 1 |
| a |
| 1 |
| a |
| 1 |
| a1−a |
| 1 |
| a |
| 1 |
| a |
| 1 |
| a |
| n |
| a |
| 1 |
| an−a |
| n |
| a |
| 1 |
| n |