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已知{An}是各项都为正数的等比数列,Sn是{An}的前n...

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...

由题意可设{an}的公比为q,q>0∵5S2=S4,∴5(a1+a2)=a1+a2+a3+a4,∴4(a1+a2)=a3+a4,即4(a1+a2)=(a1+a2)q2,∴q2,=4,解得q=2,∴a5=a1q4=1×24=16故选:C

(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...

(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×...

当q=1时,Sn+1=(n+1)a1,Sn=na1,所以limn→∞Sn+1Sn=limn→∞n+1n=1成立,当q≠1时,Sn=a1(1?qn)1?q,所以limn→∞Sn+1Sn=limn→∞1?qn+11?qn,可以看出当0<q<1时,limn→∞1?qn+11?qn=1成立,故q的取值范围是(0,1].故选B.

(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(...

设等比数列{an}的公比为q,(q>0)由题意可得2×12a3=3a1+2a2,即a1q2=3a1+2a1q,即q2-2q-3=0解之可得q=3,或q=-1(舍去)故S11?S9S7?S5=a10+a11a6+a7=a6q4+a7q4a6+a7=q4=81故答案为:81

(I)设公比为q,则a1>0,q>0,依题意:a1+a2=3(1a1+1a2)a3+a3+a5=272(1a3+1a4+1a5)即a1+a1q=3(1a1+1a1q)a1q2+a1q3+a1q4=36(1a1q2+1a1q3+1a1q4)…(2分)即a12q=3a12q6=36…..(3分)∵a1>0,q>0,故q=3,a1=1…..(5分)∴an=3n?1…..(6...

(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)∵S4=S3+8,∴a4=S4-S3=8,又∵a3=4,∴q=a4a3=84=2,a1=a3q2=1∴等比数列{an}的通项公式an=2n-1.(2)由(1)知:数列{an}的首项为1,公比为2,an=2n-1,bn=n?2n-1,∴bn=n?2n-1,∴Tn=b1+b2+b3…+bn=1+2?2+3?22+…+n?2n-12Tn=2+2?22+3?23+…+...

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