2S
(1)∵a1,a2,a7成等比数列, ∴a22=a1•a7,即(a1+d)2=a1(a1+6d), 又a1=1,d≠0,∴d=4. ∴Sn=na1+ n(n−1) 2d=n+2n(n-1)=2n2-n. (2)证明:由(1)知bn= 2Sn 2n−1= 2n(2n−1) 2n−1=2n, ∴{bn}是首项为2,公差为2的等差数列, ∴Tn= n(2+2n) 2=n2+n, ∴2Tn-9bn-1+18=2n2+2n-18(n-1)+18 =2n2-16n+36=2(n2-8n+16)+4=2(n-4)2+4≥4,当且仅当n=4时取等号.①
64bn (n+9)bn+1= 64×2n (n+9)×2(n+1)= 64n n2+10n+9= 64 n+ 9 n+10≤ 64 6+10=4. 当且仅当n= 9 n即n=3时,取等号.② ∵①②中等号不能同时取到,∴2Tn−9bn−1+18> 64bn (n+9)bn+1(n>1).
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