www.1862.net > 1.在等比数列{An}中: (1)已知A1=%1,A4=64,求q与...

1.在等比数列{An}中: (1)已知A1=%1,A4=64,求q与...

(1)a4=a1*q^3 64=(-1)*q^3 q^3=-64=(-4)^3 q=-4 S4=a1*(q^4-1)/(q-1) =(-1)*[(-4)^4-1]/(-4-1) =51 (2)a3=a1q^2=3/2 (1) s3=a1+a1q+a1q^2=9/2 (2) 由 (1)得:a1=3/(2q^2) [3/(2q^2)](1+q)=9/2-3/2=3 -2q^2+q+1=0 q=1,或q=-1/2 代入得a1=3/2...

a4=a1*q^3 64=-q^3 q=-4 S4=-1+4-16+64=51 抱歉图片拍不了

(1) ∵a4=a1q³ ∴q³=a4/a1=24/3=8 即q=2 (2) a2=a1q=3x2=6 a3=a2q=6x2=12 S4=a1+a2+a3+a4 =3+6+12+24 =45 行家正解,不明白可以追问!祝您学习进步 满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢

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(1)解:设公差为d,公比为q,则,∵a1=b1=1,a2=b2+1,a4=b4+1,∴1+d=q+1,1+3d=q3+1,∴d=q=±3,∴an=1±3(n-1),bn=(±3)n?1;(2)证明:由(1)知an=1+3(n-1),bn=(3)n?1∵cn=λ?bn+1-Sn,∴cn+1-cn=λ(bn+2-bn+1)-an+1=(3-1)λ?(3)n-1-3n>0

(1)设某等差数列为{bn},则b5=a2=a1q=64q,b3=a3=64q2,b2=a4=64q3.∵b5=b2+3(b3-b2),∴64q=64q3+3(64q2-64q3),化为2q2-3q+1=0,q≠1,解得q=12.∴an=a1qn?1=64×(12)n?1=(12)5+n.(2)∵bn=log2an=log22?5?n=-n-5.∴|bn|=n+5∴数列{|bn|...

(Ⅰ)由条件知a2-a3=2(a3-a4).(2分)即a1q-a1q2=2(a1q2-a1q3),又a1?q≠0.∴1-q=2(q-q2)=2q(1-q),又q≠1.∴q=12.(4分)∴an=64?(12)n?1=(12)n-7.(6分)(Ⅱ)bn=log2an=7-n.{bn}前n项和Sn=n(13?n)2.∴当1≤n≤7时,bn≥0,∴Tn=Sn...

(1)在等比数列{an}中,由已知可得:a1?a1q?a1q2=27a1q+a1q3=30…(3分)解得:a1=1q=3或a1=?1q=?3…(6分)(2)∵S n=a1(1?qn)1?q∴当

(1)设等比数列{an}的公比为q,由a1=1,a4=a1q3,得q=4.∴an=4n-1.∵数列{bn}的前n项和Sn满足Sn=3n2+n2,∴数列{bn}为等差数列,a1=2,a1+a2=7,∴公差d=3.∴bn=2+(n-1)×3=3n-1.(2)由(1)可得:anbn=(3n-1)?4n-1.∴Tn=2×1+5×4+8×42+…+(...

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