并且對(duì)于任意n∈N*,都有.a(chǎn)n+1=
an |
2an+1 |
∴
1 |
a1 |
1 |
an+1 |
2an+1 |
an |
1 |
an |
∴{
1 |
an |
∴
1 |
an |
∴an=
1 |
2n?1 |
(2)∵anan+1=
1 |
2n?1 |
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 |
=
n |
2n+1 |
an |
2an+1 |
1 |
an |
an |
2an+1 |
1 |
a1 |
1 |
an+1 |
2an+1 |
an |
1 |
an |
1 |
an |
1 |
an |
1 |
2n?1 |
1 |
2n?1 |
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 |
n |
2n+1 |