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已知各项均为正数的等比数列{An},首项A1=12,前n...

:(Ⅰ)设正项等比数列{an}(n∈N*)的公比为q(q>0),又a1=12,∴an=12?qn-1,∵S3+a3、S5+a5、S4+a4成等差数列,∴2(S5+a5)=(S3+a3)+(S4+a4),即2(a1+a2+a3+a4+2a5)=(a1+a2+2a3)+(a1+a2+a3+2a4),化简得4a5=a3,∴4a1q4=a1q2,化为4q...

(1)∵5S1,S3,3S2成等差数列,∴2S3=5S1+3S2…(1分)即2(a1+a1q+a1q2)=5a1+3(a1+a1q),化简得 2q2-q-6=0…(2分)解得:q=2或q=-32…(3分)因为数列{an}的各项均为正数,所以q=-32不合题意…(4分)所以{an}的通项公式为:an=2n.…(5分)(2...

(1) S4/S2=1+q²=30/6 q²=4 数列各项均为正,q>0 q=2 a1=S2/(1+q)=6/(1+2)=2 an=a1qⁿ⁻¹=2·2ⁿ⁻¹=2ⁿ 数列{an}的通项公式为an=2ⁿ (2) bn=1/[log2(an)log2(a(n+2)) =1/[log2(2ⁿ)log2(...

∵等比数列{an}的首项为2,公比为3,前n项和为Sn,∴an=2?3n-1;Sn=3n-1,∵log3[12an?(S4m+1)]=9,∴(n-1)+4m=9,∴n+4m=10,∴1n+4m=110(n+4m)(1n+4m)=110(17+4nm+4mn)≥110(17+8)=2.5当且仅当m=n=2时取等号,∴1n+4m的最小值是2.5.故答...

(Ⅰ)∵a1=3,S3=39,∴q≠1,3(1?q3)1?q=39,∴1+q+q2=13.∴q=3,或q=-4(舍),故an=3n.…(6分)(Ⅱ)∵an=3n,则an+1=3n+1,由题知:an+1=an+(n+1)dn,则dn=2×3nn+1.由上知:1dn=n+12×3n,所以Tn=1d1+1d2+…+1dn=22×3+32×32+…+n+12×3n,...

(Ⅰ)设等比数列{an}的公比是q,依题意 q>0. 由S3=14,得 a1(1+q+q2)=14,整理得 q2+q-6=0. 解得 q=2,舍去q=-3. 所以数列{an}的通项公式为an=a1?qn?1=2n. (Ⅱ)由bn=n?an=n?2n,得 Tn=1×2+2×22+3×23+…+n?2n,所以 2Tn=1×22+2×23+3×...

(1)若q=1,5S2=10a1,4S4=16a1,不满足5S2=4S4,故q≠1…(2分)由5S2=4S4得5a1(1?q2)1?q=4a1(1?q4)1?q,1+q2=54,q2=14,∵an>0,∴q=12…(5分)(2)假设满足条件的等比数列{bn}存在.由(1)得Sn=a1[1?(12)n]1?12=2a1[1?(12)n],∴bn=12...

(1)∵a4+S4,a5+S5,a6+S6成等差数列,∴2(a5+S5)=(a4+S4)+(a6+S6)…(2分)即2a6-3a5+a4=0,∴2q2-3q+1=0,∵q≠1,∴q=12,…(4分)所以等比数列{an}的通项公式为an=12n;…(6分)(2)bn=an+an+12?3n=34?(32)n,…(9分)∴数列{bn}为等比...

(Ⅰ)由S2=a1+a2=3+a2,b2=b1q=q,且b2+S2=12,S2=b2q.∴q+3+a2=12,3+a2=q2,消去a2得:q2+q-12=0,解得q=3或q=-4(舍),∴a2=q2-3=6,则d=a2-a1=6-3=3,从而an=a1+(n-1)d=3+3(n-1)=3n,bn=3n-1;(Ⅱ)∵an=3n,bn=3n-1,∴cn=3bn-λ?2an3=3n-...

(1)∵{bn}是等比数列,首项为4,公比为2, ∴bn=4?2n-1=2n+1, ∵数列{an}是等差数列,且对任意的n∈N*,都有a1b1+a2b2+a3b3+…+anbn=n?2n+3, ∴a1b1=24,∴a1=24 b1 =24 4 =4, a1b1+a2b2=2?25, ∴a2b2=2?25?24=48, ∴a2=48 b2 =48 23 =6, ∴...

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