www.1862.net > 等比数列{An}的各项均为正数,且A4A6=9,则log3A1+...

等比数列{An}的各项均为正数,且A4A6=9,则log3A1+...

由等比数列的性质得,a4a7=a5a6,因为a4a7+a5a6=18,所以a5a6=9,所以log3a1+log3a2+…+log3a10=log(a1a2a3…a10)3=log(a5a6)53=log(9)53=10,故选:A.

由a1+a2=4a3^2=16*a2*a6得 a1(1+q)=4(a1q^2)^2=16a1^2*q^6, 由后者,q^2=1/4,q>0, q=1/2.代入前者, 3a1/2=a1^2/4,a1>0, a1=6. (1)an=6*(1/2)^(n-1)=3/2^(n-2). (2)bn=log3-(n-2)=2+log3-n, 1/[bnb]=1/b-1/bn, ∴Tn=1/b2-1/b1+1/b3-1/b2+……+1/b-1...

等比数列An 设等比为Q 那么AN=A1*Q^(N-1) A3=A1*Q^2 A9=A1*A^8 A1*Q^2*A1*Q^*=A1^2*Q10=9=(A1*Q^5)^2=9 log3A1+log3A2+...+log3A11=LOG3 (A1*A2*...*A11) 又A1*A2*...*A11=A1*A1*Q*A1*Q^2*.......A1*Q^10=A1^11*Q^(1+10)*5)=A1^11*Q^55=(A1*Q^5)...

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...

设等比数列{an}的公比为q,∵a1=3,a4=81,∴81=3×q3,解得q=3.∴an=3n.∴bn=log3an=log33n=n.∴1bnbn+1=1n(n+1)=1n?1n+1.∴Sn=(1?12)+(12?13)+…+(1n?1n+1)=1?1n+1=nn+1.∴S2013=20132014.故答案为20132014.

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

解: 由于{an}为等比数列 则:a5a6=a4a7=a3a8=a2a9=a1a10 又a5a6+a4a7=18 则: 2a5a6=18 a5a6=9 则: log3(a1)+log3(a2)+...+log3(a9)+log3(a10) =log3[a1*a2*a3*...*a10] =log3[(a1a10)*(a2a9)*...*(a5a6)] =log3[9*9*...*9] =log3[9^5] =log3[3^10...

log3a1+log3a2+…+log3a10=log3(a1a10)+log3(a2a9)+…log3(a5a6)=5log3(a5a6)=10故答案为:10

(Ⅰ)设数列{an}的公比为q,由a32=9a2a6得a32=9a42,所以q2=19.由条件可知各项均为正数,故q=13.由2a1+3a2=1得2a1+3a1q=1,所以a1=13.故数列{an}的通项式为an=13n.(Ⅱ)bn=loga13+loga23+…+logan3=-(1+2+…+n)=-n(n+1)2,故1bn=-2n(n+1)=-2...

∵a10a11=a9a12,∴a10a11+a9a12=2a10a11=486∴a10a11=486∴log3a1+log3a2+…log3a20=log3(a10a11)10=10log3(a10a11)=50.故答案为:50.

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