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共回答了2680个问题采纳率:81.2% 举报
(I)n=1时,6a1=a12+3a1+2,且a1>1,解得a1=2. n≥2时,6Sn=an2+3an+2,6Sn-1=an-12+3an-1+2,两式相减得(an+an-1)(an-an-1-3)=0, ∵an+an-1>0, ∴an-an-1=3, ∴{an}为等差数列, ∵a1=2, ∴an=3n-1. (II)证明:∵数列{bn}满足an(2bn−1)=1, ∴bn=log2 3n 3n−1 ∴Tn=b1+b2+…+bn=log2( 3 2 × 6 5 ×…× 3n 3n−1 ) 要证2Tn+1<log2(an+3),即证2log2( 3 2 × 6 5 ×…× 3n 3n−1 )+1<log2(an+3) 即证( 3 2 × 6 5 ×…× 3n 3n−1 )2< 3n+2 2 即证 2( 3 2 × 6 5 ×…× 3n 3n−1 )2 3n+2 <1 令cn= 2( 3 2 × 6 5 ×…× 3n 3n−1 )2 3n+2 , ∴ cn+1 cn = 9n2+18n+9 9n2+21n+10 <1 ∵cn>0,∴cn+1<cn, ∴{cn}是单调递减数列 ∴cn≤c1= 2×( 3 2 )2 3×1+2 = 9 10 <1 ∴cn= 2( 3 2 × 6 5 ×…× 3n 3n−1 )2 3n+2 <1 故2Tn+1<log2(an+3).
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