解(1)当n=1时,由题意可得6a1=a12+3a1+2
∴a1=1或a1=2
当n≥2时,6Sn=an2+3an+2,6Sn-1=an-12+3an-1+2,
两式相减可得(an+an-1)(an-an-1-3)=0
由题意可得,an+an-1>0
∴an-an-1=3
当a1=1时,an=3n-2,此时a42=a2?a9成立
当a1=2时,an=3n-1,此时a42=a2?a9不成立
故an=3n-2,
(2)∵bn=23n-2,是以公比q=8的等比数列,
∴数列的前n项和为Tn=
=2(1?8n) 1?8
(8n?1)2 7