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已知等比数列{An}各项均为正数,数列{Bn}满足Bn=log...

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

解答:(1)解:设数列{an}的公比为q(q>0),由a1+a3=10,a3+a5=40,则a1+a1q2=10 ①a1q2+a1q4=40②,∵a1≠0,②÷①得:q2=±2,又q>0,∴q=2.把q=2代入①得,a1=2.∴an=a1qn?1=2×2n?1=2n,则bn=log2an=log22n=n;(2)证明:∵c1=1<3,cn+1-cn=...

(1)设公比为q(q≠1),a3=a1q2,a5=a1q4 …(2分)代入:(bn+1-bn+2)?log2a1+(bn+2-bn)?log2a3+(bn-bn+1)?log2a5=0得∴[(bn+1-bn+2)+(bn+2-bn)+(bn-bn+1)]log2a1+2[(bn+2-bn)+2(bn-bn+1)]log2q=0即(bn+2+bn-2bn+1)log2q=0∵q≠1...

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a3a5=a42=4,又等比数列{an}的各项均为正数,∴a4=2,则数列{log2an}的前7项和S7=loga12+loga22+…+loga72=log(a1 ?a7 )(a2?a6)(a3?a5) a42=loga472=log272=7.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...

由等比数列的性质可知a2a4=a32=4,又∵an>0∴a3=2即a1q2=2①∵S3=a1(1?q3)1?q=14,即a1(1+q+q2)=14②②÷①可得q2+q+1q2=7解方程可得q=12或q=?13(舍)∴a1=8,an=8?12n?1=12n?4∴bn=log2an=4-n∴S6=3+2+1+0-1-2=3故选C

(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.

an=4^n bn=log4(4^3n)-2=3n-2

(1) an=a1q^(n-1) bn=loga1+(n-1)log(q)为等差数列 (2) b3=6/3=2 b1b5=0 b1=loga1>0 所以b5=0, b1=4 所以bn=5-n Sn=5n-1/2n^2+1/2n=-1/2n^2+9/2n an=2^(5-n)

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