S
(Ⅰ)b1+b3+b5=log2(a1a3a5)=log2(a13q6)=6⇒a13q6=26⇒a1q2=4, ∵a1>1,∴b1=log2a1≠0, 又b1b3b5=0,若b3=0,则log2a3=log2(a1q2)=0,即a1q2=0,这与a1q2=4矛盾, 故b5=log2(a1q4)=0⇒a1q4=1. ∴q2= 1 4,q= 1 2,a1=16. ∴an=16•( 1 2)n−1=25-n. (Ⅱ)∵bn=log2an=log225−n=5-n,∴{bn}是首项为4,公差为-1的等差数列, ∴Sn= 9n−n2 2, Sn n= 9−n 2. 故{ Sn n}是首项为4,公差为- 1 2的等差数列.∵n≤8时, Sn n>0; n=9时, Sn n=0; n>9时, Sn n<0.故当n=8或n=9时, S1 1+ S2 2+…+ Sn n最大.
已知等比数列an的首项a1>1,公比q>0,设bn=log2an,(n属于N*)且b1+b2+b3=6,b1b2b3=0
已知等比数列(An)的首项a1>1公比q>0设Bn=log2an且b1+b2+b3=6,b1.b2.b3=0记(Bn)的
在等比数列an中,a1>1,公比q>0,设bn=log2an,且b3=2,b5=0
已知等比数列an的首项a1>1,公比q>0,设bn=log2a2,且b1+b3+b5=6,b1b3b5=0,又bn的前n
已知数列{an}是等比数列,首项a1=8,公比q>0,令bn=log2an,设sn为{bn}的前n项和,若
设等差数列an的前n项和为Sn,等比数列bn的前n项和为Tn,已知数列bn的公比为q(q>0),a1=b1=1,S5=4
设等差数列{an}的前n项和为Sn,等比数列{bn}的前n项和为Tn,已知数列{bn}的公比为q(q>0),a1=b1=
已知等比数列{an}的首项a1>0,公比q>0.设数列{bn}的通项bn=a(n+1)+a(n+2),数列{an},{b
已知数列{an}的前n项和Sn=n2(n∈N*),数列{bn}为等比数列,且满足b1=a1,2b3=b4
设等比数列{an}的前n项和为Sn 等比数列{bn}的前n项和Tn 已知数列{bn}的公比q>0 a1=b1=1 S5=
已知等比数列an中,a1=2,a4=16,数列bn中,b1=1且bn-bn-1=log2an(n≥2),求bn
已知数列[an]为等差数列,公差d≠0;[bn]为等比数列,公比为q,若a1=b1,a3=b3,a7=b5,且an=bm
|