∴Sn+1=2an+1-3n-3,
两式相减,得a n+1=2an+1-2an-3,即an+1=2an+3,
∴an+1+3=2(an+3),
所以数列{bn}是以2为公比的等比数列,
由已知条件得:S1=2a1-3,a1=3.
∴首项b1=a1+3=6,公比q=2,
∴an=6•2n-1-3=3•2n-3.
(2)∵nan=3×n•2n-3n
∴Sn=3(1•2+2•22+3•23+…+n•2n)-3(1+2+3+…+n),
2Sn=3(1•22+2•23+3•24+…+n•2n+1)-6(1+2+3+…+n),
∴-Sn=3(2+22+23+…+2n-n•2n+1)+3(1+2+3+…+n)
=3×
2(2n-1) |
2-1 |
3n(n+1) |
2 |
∴Sn=(6n-6)•2n+6-
3n(n+1) |
2 |