www.1862.net > 等比数列{An}的前n项和为sn,已知s4=1,s8=17,求{An}...

等比数列{An}的前n项和为sn,已知s4=1,s8=17,求{An}...

设等比数列{an}的公比为q,由S4=1,S8=17,得S8-S4=17-1=16,∴q4=S8?S4S4=16,解得q=±2,当q=2时,由S4=a1(1?q4)1?q=15a1=1可得a1=115,∴Sn=a1(1?qn)1?q=115(2n-1);当q=-2时,由S4=a1(1?q4)1?q=-5a1=1可得a1=?15,∴Sn=a1(1?qn)1?q=115[(-2)...

(1)∵等差数列{an},其前n项和为Sn,若S4=4S2,a2n=2an+1,∴4a1-2d=0,a1=d-1,∴a1=1,d=2,∴an=2n-1(2)∵an=2n-1,∴2n-1>2m,2n-1<22m,∴2m-1+12<n<22m-1+12,即项数22m-1-2m-1,∴①bm=22m?1?2m?1∵cm=222m?1?bm,∴Cm=22m?1,∴c1=2,Cn+1C...

(1)解:设数列{an}的公比为q,∵且S3,S2,S4成等差数列,∴S3+S4=2S2,即(a1+a2+a3)+(a1+a2+a3+a4)=2(a1+a2)∴2a3+a4=0,q=a4a3=-2,∴an=a1qn-1=(-2)n-1;(2)证明:|an|=2n-1,bn=log22n?1=n?1,∴bn+1|an|=n2n?1,∴Tn=120+221+322...

s3=?

(1)∵a4+S4,a5+S5,a6+S6成等差数列,∴a5+S5-(a4+S4)=a6+S6-(a5+S5),∴2a5-a4=2a6-a5,∴2a6-3a5+a4=0.∵数列{an}为等比数列,∴a5=a4q,a6=2a4q2,∴2q2-3q+1=0,∴(2q-1)(q-1)=0.∵数列{an}公比不为1,∴q=12.∴an=2×(12)n?1,∴an=(12)...

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)∵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)由题意得a1+2d=3,4a1+6d=10,解得a1=1,d=1,从而数列{an}的通项公式为an=n.(Ⅱ)bn=1anan+1=1n(n+1)=1n?1n+1,∴Tn=(1-12)+(12?13)+…+(1n?1n+1)=nn+1.

①因为{an}为等比数列,Sn是其前n项和,在若S4,S8-S4,S12-S8求和为0时,则就不成等比数列;②由题意画出图象为:易有当x∈(-π2,π2)时,y=sinx与y=tanx的图象交点只有一个为(0,0),所以②正确;③有空间想象出图象为:在正四棱锥中,点M为边CD...

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