数列bn各项均为正数,若b3=1,bn²=4b²n 1,则通项

∵等比数列{an}的各项均为不等于1的正数，数列{bn}满足bn=lnan，b3=18，b6=12，∴a3=a1q2=eb3=e18，a6＝a1q5=eb6=e12，∴a6a3=q3=e12e18=e

因为a2+a4=2a3,b2*b4=(b3)²所以2a3=b3,(b3)²=a3那么(b3)²=1/2*a3而b3>0,所以b3=1/2于是a3=1/4那么公差d=(1/

a(n+1)=√(bn*b(n+1))2bn=√(bn*b(n-1))+√(bn*b(n+1))2√bn=√b(n-1)+√b(n+1)得证第二问√b1=√2b2=a2^2/b1=4.5√b2=√(9

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*b

an=(n-1)d+1bn=q^(n-1)2d+1+q^4=214d+1+q^2=132*q^4-q^2=28(2q^2+7)(q^2-4)=0q^2=4因为q大于零,所以q=2,d=2an=2n-1

n=1,时a1=S1=1n>=2时,an=Sn-S(n-1)=n^2-(n-1)^2=2n-1故an=2n-1a5=9b1=a1=1a5*b3=1,b3=1/9q^2=b3/b1=1/9q=1/3故b

(1)C(2)arccos1/3

∵数列Bn满足Bn=lgAn又∵B3=18,B6=12∴A3=10^18,A6=10^12又∵等比数列An的各项均为不等于1的正数∴A6=A3*q^3即q=10^(-2)∴A1=A3/q^2=10^2

132解；bn=lgan,所以an=10^bn,因为{an}为等比数列,b3=18,b6=12,代入an=10^bn,得a3=10^18,a6=10^12,用a6/a3,得公比q^3=1/(10^6)

(1)a3=a1*q^2=e^(b2)=e^18a6=a1*q^5=e^(b6)=e^12则:a6/a3=q^3=e^12/e^18=e^(-6)得:q=e^(-2),a1=e^22等比数列{an}的

An满足A2=2,A5=8a5-a2=3d=6d=2a1=a2-d=0an=2(n-1)b3=a3=4T3=7=b1+b2+b3b1+b2=3b2^2=b1b3=4b1=4(3-b2)b2=2所以公比

楼主您好：An满足A2=2,A5=8a5-a2=3d=6d=2a1=a2-d=0an=2(n-1)b3=a3=4T3=7=b1+b2+b3b1+b2=3b2^2=b1b3=4b1=4(3-b2)b2=

a(n+1)=√[bn*b(n+1)]2bn=an+an+12bn=√[bn*b(n-1)]+√[bn*b(n+1)]2√bn=√b(n-1)+√b(n+1)所以数列{√bn}为等差数列√b1=√2(

(1).a3+a5=2a4=21a4=21/2d1=(a4-a1)/3=19/6a1=1an=1+(n-1)*19/6b5=13-a4-d1=q2=4根号b5bn=(1-q2的n次方）/(1-q2)具

由于:5^[an],5^[bn],5^[a(n+1)]成等比数列则有:{5^[bn]}^2=5^[an]*5^[a(n+1)]5^[bn^2]=5^[an+a(n+1)]则:2bn=an+a(n+1)

靠错位相减法怎么打啊太复杂了·····设等差数列的公差为d等比数列的公比为q由题意得1+2d+q^4=21（1）1+4d+q^2=13（2）（1）*2-（2）得2q^4-q^2-28=0解得q^2=4

an,bn,an+1成等差数列2bn=an+a(n+1)bn,an+1,bn+1成等比数列[a(n+1)]^2=bn*b(n+1)根据上述2式得2bn=根号（b(n-1)*bn）+根号(bnb(n+1

a1,qb1=log2a1b2=log2a2=loga1+log2qb3=log2a3=log2a1q^2=log2a1+2log2q相加得log2a1q=log2a2=1a2=a1q=2log2a1

题目说清楚,不知道让求什么,条件也不明确再问：上面那个是网址，请您去那里看一下，谢谢！再答：条件是吗再问：对呀A任意n属于N+,an>bn,则an+1>bn+1B存在m属于N+,任意n>m,an=bn

log2A(n＋1)=log2An＋1=log2[2An],则：A(n＋1)=2An,则[A(n＋1)]/[An]=2=常数,则数列{An}是以A1=1为首项、以q=2为公比的等比数列,得：An=2^