在等比数列an中,q=4,且a1a2a3
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![在等比数列an中,q=4,且a1a2a3](/uploads/image/f/3263184-0-4.jpg?t=%E5%9C%A8%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97an%E4%B8%AD%2Cq%3D4%2C%E4%B8%94a1a2a3)
a1*a2*a3*a4*.*a29*a30=2^30等价于(a1*a30)的15次方=2^30推得(a1*a30)的5次方=2^10同理a3*a6*a9*...*a30等价(a3*a30)的5次方=(
n=a(n+1)-an=a1q^n-a1q^n-1=a1q^n(1-1/q)b(n+1)=a1q^(n+1)(1-1/q)b(n+1)/bn=q(定值)所以{bn}是等比数列b1=1*q(1-1/q)
a2*a(n-1)=a1*an=128而a1+an=66解得a1=2an=64或a1=64an=2当a1=2an=64时由求和公式有126=(2-64q)/(1-q)得q=2n=6当a1=64an=2
设A1A2=a则:由于在数列{An}中An小于0故a>0,且An+1An+2/AnAn+1>0即q>0;由题中:2AnAn+1+An+1An+2>An+2An+3得2aq^(n-1)+aq^n>aq^
a3=a1*q^2,^2表示平方a5=a1*q^4...a1+a3+a5+...+a99=a1(1+q^2+...+q^98)=a1(1-q^50)/(1-q^2)=a1(1-q^50)/[1-q][
a(n+2)=a(n)+a(n+1)a(n)*q^2=a(n)+a(n)*qq^2=1+q利用求根公式,并且q>0.可知,q=(1+√5)/2
a1+an=a1+a1q^(n-1)=66a1q^(n-1)=66-a1a2*a(n-1)=a1q*a1*q^(n-2)=a1*a1*q^(n-1)=128a1(66-a1)=128a1^2-66a1
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a1=1bn=a1*q^n-a1*q^(n-1)=q*q^(n-1)-q^(n-1)=(q-1)*q^(n-1)b(n+1)=(q-1)*q^nq-1≠0所以b(n+1)/bn=q所以是等比数列
这个图片不知道行不行啊再问:{an+1}为等比数列怎麽会有An+1+An-1=An再答:这是按照上面的公式算出来的啊,是等于2An因为an是等比数列,所以an+1*an-1=an*an
a1+an=66a2*an-1=a1*an=128所以可得:a1=2,an=64或a1=64,an=2当a1=2,an=64时有:Sn=a1(1-q^n)/(1-q)=(a1-anq)/(1-q)=(
a2an-1=a1anai,an是方程x^2-66x+128=0的根a1=2,an=64,sn=a1(1-q^n)/(1-q)=(a1-qan)/(1-q)=128解q=2,2×2^(n-1)=2^6
∵等比数列{an}中,an>0,且an+2=an+an+1,∴a1qn+1=a1qn-1+a1qn,∴q2=1+q,解得q=1±52,又∵q>0.∴q=1+52.故答案为1+52.
解题思路:利用等比数列的知识求解。解题过程:见附件最终答案:略
设an=a1×q^(n-1)an+2=an+a(n+1)a1×q^(n+1)=a1×q^(n-1)+a1×q^nq^2=1+qq=(1±√5)/2再问:q^2=1+q这部是什么意思再答:a1×q^(n
可以知道an=a1*2^(n-1);由log2a1+log2a2+log2a3+……+log2a10=25;得log2a1+log2a1*2^1+log2a1*2^2+……+log2b*2^9=25;
a3=a1*q^2,^2表示平方a5=a1*q^4...a1+a3+a5+...+a99=a1(1+q^2+...+q^98)=a1(1-q^50)/(1-q^2)=a1(1-q^50)/[1-q][
等比数列an中a1=1/2,a4=4则公比q=(a4/a1)开3次方=8开3次方=2a1+a2+…+an=Sn=a1(1-q^n)/(1-q)=1/2(1-2^n)/(1-2)=2^(n-1)-1/2
设等差数列{an}的公差为d,∵a1,a3,a4成等比数列,∴(a1+2d)2=a1(a1+3d),解得d=0或d=-14a1,当d=0时,公比q=1,当d=-14a1时,公比q=a3a1=14,故答