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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...

因为|x|-1能取遍一切正实数,所以函数y=log2(|x|-1)值域是R,故(1)正确;若{an}是公比为-1的等比数列,当Sk=0时则Sk,S2k-Sk,S3k-S2k(k为常数且k∈N)不是等比数列,故(2)错误;设两定点分别为A(m,n),B(c,d),设所求点为(x,y)...

由题意可得an=a1qn-1=8?qn-1,所以bn=log2an=log2(8?qn-1)=3+log2qn?1=3+(n-1)log2q,上式为关于n的一次函数的形式,故数列{bn}为等差数列,又知S3是数列{Sn}中的唯一最大项,故b3>0b4<0代入可得3+2log2q>03+3log2q<0,解得?32<log2q<...

设正项等比数列{an}的公比为q,设正项等比数列{bn}的公比为p,则数列{lgan}是等差数列,公差为lgq,{lgbn}是等差数列,公差为lgp.故 Sn =n?lga1+n(n?1)2? lgq,同理可得 Tn =n?lgb1+n(n?1)2? lgp.又SnTn=n2n+1=lga1+n?12lgqlgb1+n?12lgp,∴lo...

设等比数列{a n }的公比为q,依题意有2(a 3 +2)=a 2 +a 4 ,(1)又a 2 +a 3 +a 4 =28,将(1)代入得a 3 =8.所以a 2 +a 4 =20.于是有 a 1 q+ a 1 q 3 =20 a 1 q 2 =8 解得 a 1 =2 q=2 或 a 1 =32 q= 1 2 又{a n }是递增的,故a 1 =2,q=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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