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: 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...

:(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...

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)...

(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)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)∵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,...

(1)设等差数列{an}的公差为d,等比数列{bn}的公比为q,则由题意知b1q(2a1+d)=4b1q2(3a1+3d)=154,把a1=3,b1=1代入上式,解得d=2q=12或d=?65q=56,∵等差数列{an}的各项均为正数,∴舍去d=-65,∴an=3+(n-1)×2=2n+1,bn=1×(12)n?1=(12)...

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