www.1862.net > 已知等比数列{An}的各项均为正数,其前n项和为Sn....

已知等比数列{An}的各项均为正数,其前n项和为Sn....

(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(...

(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×...

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...

(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...

解: 数列是等比数列,S2,S4-S2,S6-S4成等比数列,公比为q² 由S4=2S2+1得S4-S2=S2+1 S4-S2=q²S2代入,得q²S2=S2+1 S2=1/(q²-1) (S4-S2)²=S2·(S6-S4) S6=(S4-S2)²/S2 +S4 =(q²S2)²/S2 +2S2+1 =q⁴...

S1/T1=(3+1)/4=1, 即S1=T1, 得a1=b1 令an公比为q, bn公比为p S2/T2=(a1+a1q)/(a1+a1p)=(1+q)/(1+p)=(3²+1)/4=5/2, 得:2+2q=5+5p, 即q=(3+5p)/2 S3/T3=(3³+1)/4=7=(1+q+q²)/(1+p+p²) 代入q,化简得: 7+7p+7p²=1+(3+5p)...

(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-...

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.

:(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...

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

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