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已知等比数列{An}满足An>0,n=1,2,…,且A5?A2n%...

在等比数列{an}中,a5?a2n-5=22n(n≥3),即a5?a2n-5=a2n=(2n)2,(n≥3),∴an=2n,(n≥3),∴a1=2,a2=4.∴log2an=log22n=n,∵n≥1时,log2a1+log2a2+log2a3+…log2an,∴log2a1+log2a2+log2a3+…log2an=1+2+…+n=n(n+1)2,故选:B.

由题意等比数列{an}a>0,n=1,2,…,当n>1时,log2a1+log2a3+log2a5+…+log2a2n+1=log2a1a3a5…a2n+1.又a5?a2n-5=22n(n≥3)∴a1a3a5…a2n+1=2(n+1)2∴log2a1+log2a3+log2a5+…+log2a2n+1=log22(n+1)2=(n+1)2故选:C.

解:∵{An}为等比数列,∴A5×A2n-5=A1×A2n-1=A2×A2n-2=……=2^2n。而㏒2A1+㏒2A2+㏒2A3+……+㏒2A2n-1=㏒2(A1×A2×A3×……×A2n-2×A2n-1)=㏒2{2^2n}^(n/2)=㏒2(2^n^2)=n^2。考查等比数列的性质及对数运算,如有问题,敬请追问,万望采纳!祝您学习...

(Ⅰ)设等差数列的公差为d,等比数列的公比为q,则a1=1,a2=2,a3=1+d,a4=2q,a5=1+2d,∵S3=a4,a3+a5=a4+2.∴4+d=2q(1+d)+(1+2d)=2+2q,解得d=2,q=3,∴an=n,n=2k?12?3n2?1,n=2k.(Ⅱ)当n=1时,S2n=2n+n2.当n≥2时,S2n=(1+2n?1)n2+2...

(1)当n=1时,4a1=a22?5,a22=4a1+5,∵an>0∴a2=4a1+5(2)当n≥2时,满足4Sn=a2n+1?4n?1,n∈N*,且4Sn?1=a2n?4(n?1)?1,∴4an=4Sn?4Sn?1=a2n+1?a2n?4,∴a2n+1=a2n+4an+4=(an+2)2,∵an>0,∴an+1=an+2,∴当n≥2时,{an}是公差d=2的等差数...

(I)∵数列{an}满足:a1=1,an+1=an+1n为奇数2ann为偶数(n∈N*),设bn=a2n-1,∴b2=a3=2a2=2(a1+1)=4,b3=a5=2a4=2(a3+1)=10,同理,bn+1=a2n+1=2a2n=2(a2n+1+1)=2(bn+1)=2bn+2.(II)①b1=a1=1,b1+2≠0,bn+1+2bn+2=2bn+2+2bn+2=2,∴...

(1)设{an}的公比为q,∵a3=1,∴a4=q,a5=q2,a6=q3.∵a4,a5+1,a6成等差数列,∴2(q2+1)=q+q3,解得q=2. (2分)∴an=a3qn-3=2n-3. (3分)当n=1时,a1b1=S1=1,∴b1=a1=14.(4分)当n≥2时,anbn=Sn?Sn?1=n?2n?3,∴bn=14 n=11n n≥2...

(1)∵{bn}是等比数列,首项为4,公比为2,∴bn=4?2n-1=2n+1,∵数列{an}是等差数列,且对任意的n∈N*,都有a1b1+a2b2+a3b3+…+anbn=n?2n+3,∴a1b1=24,∴a1=24b1=244=4,a1b1+a2b2=2?25,∴a2b2=2?25?24=48,∴a2=48b2=4823=6,∴d=a2-a1=6-4=2,...

数列{an}是等比数列,设其公比为q,①∵a2n+2a2n=a1q2n+1a1q2n?1=q2为常数,∴数列{a2n}是等比数列;②∵an+1+anan+an?1=q(an+an?1)an+an?1=q为常数,∴数列{an+an-1}是等比数列;③若数列{an}是常数列,且各项为1,则数列{lgan}不是等比数列;④∵|a...

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