www.1862.net > 已知等比数列{An}各项均为正数,数列{Bn}满足Bn=log...

已知等比数列{An}各项均为正数,数列{Bn}满足Bn=log...

an为等比数列 由于bn=log2an,则bn为等差数列,设bn公差为d 则 b1+b2+b3=3 推出 3b1+3d=3 进而 d=1-b1 再由题:b1b2b3=-3 推出b1^3+3*d*b1^2+2*d^2*b1=-3 于是可以解得b1=-1或b1=3 若b1=-1 d=1-b1=2,b2=b1+d=1; a1=0.5,a2=2; 所以公比为4 an...

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

:(Ⅰ)因为数列{an}是各项均为正数的等比数列,且a1+a2=20,a3=64,所以a1(1+q)=20a1q2=64解得a1=4,q=4∴an=4n,bn=12log2an=n(2)∵Tn=1b1b2+1b2b3+1b3b4+…+1bnbn?1=11?2+12?3+…+1n(n?1)=1?12+12?13+…+1n?1?1n=1?1n=n?1n

(1)设等比数列的公比为q,有a1q=4a1q2+a1q3=24,解得a1=2,q=2,所以an=2n;(5分)(2)由(1)知bn=log22n=n,有an+bn=2n+n,从而Tn=(2+22+…+2n)+(1+2+…+n)=2n+1+n(n+1)2-2.(10分)

(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...

解:设{an}的公比为q,则有,a3=a1q^2=8,s2=a1+a2=a1(1+q)=48,解得q=1/2,-1/3(舍去),∴a1=32,an=32*(1/2)^(n-1)=2^(6-n)。∴bn=4log2an=4(6-n)。假设存在m∈N+,使得bm*bm+1/bm+2是数列{bn}中的第k项,则将bm等项代入、整理有k=m-1+2/(m-4)。...

(Ⅰ)∵{an}是正项等比数列,a2a8=4,∴a52=4,解得a5=2,又∵a4+a6=203,∴a1q4=2a1q3+a1q5=203,两式相除得:q1+q2=310.…(2分)解得q=3或者q=13,∵{an}为增数列,∴q=3,a1=281.…(4分)∴an=a1qn-1=281?3n-1=2?3n-5.∴bn=log3an2=n-5.…(6...

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

(1)证明:由Sn,an,12成等差数列,知2an=Sn+12,当n=1时,有2a1=a1+12,∴a1=12,当n≥2时,Sn=2an-12,Sn-1=2an-1-12,两式相减得an=2an-2an-1(n≥2),即an=2an-1,由于{an}为正项数列,∴an-1≠0,于是有anan?1=2(n≥2),∴数列{an}从第二项...

(1)a3+2是a2和a4的等差中项,且a1=2,∴2(a3+2)=a2+a4即2(a1q2+2)=a1q+a1q3∴q=2∴an=a1qn-1=2n(2)∵bn=2nlog22n=n?2n∴Tn=b1+b2+…+bn=1?21+2?22+3?23+…+n?2n∴2Tn=1?22+2?23+…+(n-1)?2n+n?2n+1∴-Tn=2+22+23+…+2n-n?2n+1∴Tn=(n-1)?2n+1+2

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