∵a1=1,an+1=
,an an+3
∴两边取倒数,得
=1+1 an+1
,3 an
令
+t=3(1 an+1
+t)即t=1 an
,1 2
∴
+1 an
=(1 2
+1 a1
)?3n-1=1 2
,3n 2
∴an=
,2
3n?1
∵bn=(3n-1)?
?ann 2n
∴bn=n?(
)n-11 2
∴Tn=1?(
)0+2?(1 2
)1+3?(1 2
)2+…+n?(1 2
)n-11 2
Tn=1?(1 2
)1+2?(1 2
)2+3?(1 2
)3…+n?(1 2
)n1 2
相减得
Tn=1+1 2