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已知等比数列{An}满足An>0,n=1,2,…,且A5?A2n%...

由a5?a2n-5=an2=22n,且an>0,解得an=2n,则log2a1+log2a2+log2a3+…+log2a2n-1=log(a1a2n?1)?(a2a2n?2) …an 2=log2n22=n2.故答案为:n2

回答:(1)n大于等于1的作用是:log2a(2n--1)有意义。 (2)题中的log2a(1)+log2a(2)+.....+log2a(2n--1)=log2(a1)*a(2)......*a(2n--1) 是运用了对数的性质“对数的和等于积的对数。”如“logaM+logaN=log a(M*N).” (3) A5*A2n--5=A1*A2n--1. 这...

在等比数列{an}中,由a3a2n?3=32n(n≥2),得:an2=a3a2n?3=32n.因为an>0,所以an=3n.则log3a1+log3a3+…log3a2n-1=log3(a1a3…a2n-1)=log331+3+…+(2n?1)=log33(1+2n?1)n2=log33n2=n2.故答案为n2.

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...

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(1)解:由a2n-1,a2n,a2n+1成等差数列,可知a1,a2,a3成等差数列,a3,a4,a5成等差数列,由a2n,a2n+1,a2n+2成等比数列,可知a2,a3,a4成等比数列,则2=a1+a3a32=a2a42a4=a3+a5,又a2=1,a5=3,∴2=a1+a3a32=a42a4=a3+3,则2a32=a3...

(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...

(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,∴...

(Ⅰ)因为a2n+1=2a2n+anan+1,所以(an+1+an)(2an-an+1)=0,因为an>0,?所以有2an-an+1=0,即2an=an+1,所以数列{an}是公比为2的等比数列,?由a2+a4=2a3+4得2a1+8a1=8a1+4,解得a1=2.从而数列{an}的通项公式为an=2n.…(6分)(II)bn=na...

(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...

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