已知数列{an}的各项均为正数,且前n项之和Sn满足6Sn=an2+3an+2,且a2,a4,a9成等比数列.(1)求数列{an

2025-05-18 11:59:38
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回答1:

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
=
2
7
(8n?1)