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在等比数列An中,An大于0,且A5A6=9,求log3A2+log...

因为a5?a6=a1q4?a1q5=a12q9,而a2?a9=a1q?a1q8=a12q9,所以a5?a6=a2?a9=9,则log3a2+log3a9=loga2a93=loga5a63=log323=2.故答案为:2

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)...

(1)设等比数列的公比为q,由a2=9,a5=243,得q3=a5a2=2439=27.∴q=3.则an=a2qn?2=9×3n?2=3n;(2)bn=log3an=log33n=n.则Tn=b1+b2+…+bn=1+2+…+n=n(n+1)2.

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...

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...

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

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.

a1,q b1=log2a1 b2=log2a2=loga1+log2q b3=log2a3=log2a1q^2=log2a1+2log2q 相加得log2a1q=log2a2=1 a2=a1q=2 log2a1=x log2a1q^2=2-x b1*b2*b3=x*1*(x-2)=-3 x=1 or x=-3 代入就是了

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...

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