设数列an是公比大于1的等比数列,sn为数列an的前n项和,已知s3=7,eg
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n≥2时,Sn=4a(n-1)+2,与S(n+1)=4an+2相减,得:a(n+1)=4an-4a(n-1),即:a(n+1)-2an=2[an-a(n-1)],则:bn=2b(n-1),其中n≥2.
Sn=2an-2n则Sn+1=2an+1-2(n+1)an+1=Sn+1-Sn=2an+1-2an-2则an+1-2an=2所以{an+1-2an}是等差数列(2)an+1-2an=2则an+1+2=
loganan+1-log(an-1)an=logan(an×q)-log(an-1)(an-1×q)=1+loganq-1-log(an-1)q=loganq-log(an-1)q<0所以递减
【参考答案】1、先求An通项公式设数列An公比为q(q>0)则S4=2S2即1+q+q²+q³=5(1+q)解得q=-1、-2或2由于q>0故q=2∴An=2^(n-1)2、再求B
设数列An的公比为q则:An=(a1)q^(n-1)而:a10^2=a15所以:((a1)q^(10-1))^2=(a1)q^(15-1)q^4=1/a1因q>1,因此:a1>0设另有数列Bn,Bn=
{1+an}的首项为3(1+an)=3*2^(n-1)1+a(6)=3*2^5=96a(6)=95
(an*an+1)/(an-1*an)=3=>an+1/an-1=3=>a2n=3^n,a2n-1=2*3^(n-1)=>bn=5*3^(n-1)
易得ana(n+1)=a1a2q^(n-1)=2q^(n-1)故2q^(n-1)+2q^n>2q^(n+1)即1+q>q^2解得(1-√5)/2再问:q>0时,求an的前2n项和sn再答:ana(n+
1=a1a2=r,故bn=r*q^(n-1)又b(n+1)/bn=a(n+1)*a(n+2)/(an*a(n+1))=a(n+2)/an、b(n+1)/bn=q可得当n为奇数时an=a1*q^((n+
由题意得1S3=a1+a2+a3=7……1;6a2=a1+1+a3+6……22式+1式得a2=2……3将3式代入12得q=2或1/2a1=4或1an=4*(1/2)^(n-1)或an=2^(n-1)2
(A10)^2=A15=A10*q^5,所以:A10=q^5=A5*q^5.,所以A5=1故A1=q^(-4),A2=q^(-3),A3=q^(-0),A4=^(q^-1).1/A1=q^4=A9,1
(n+1)=a(n+1)+1=[2an+1]+1=2an+2=2(an+1)=2bn,所以{bn}是公比为2的等比数列.b1=a1+1=2,所以bn=b1*q^(n-1)=2*2^(n-1)=2^n.
数列{Sn+1}是公比为2的等比数列S(n)+1=2^(n-1)(S1+1)=2^(n-1)(a1+1)①S(n-1)+1=2^(n-2)(a1+1)②①-②得an=2^(n-2)(a1+1),n≥2
Tn=1/a1+1/a2+……+1/anTn/q=1/a2+……+1/an+1/(q*an)Tn-Tn/q=1/a1-1/(q*an)Tn=q/a1(q-1)-1/an(q-1)
因为{Sn+1}是公比为2的等比数列,设首项为a所以Sn+1=a2^(n-1)Sn=a2^(n-1)-1n≥2时,有an=Sn-Sn-1=(a2^(n-1)-1)-[a2^(n-2)-1]=a2^(n
1.2*3a2=a1+3+a3+4(1)a1+a2+a3=7(2)a2=a1*q,a3=a1*q^2(3)三个式子连列得;a1=1,q=22.f(x)=2cosx(sinx-cosx)+1=2sinx
1、设{an}公比为qa1+a3=7-a2a1+3,3a2,a3+4构成等差数列2*3a2=a1+3+a3+46a2=7-a2+7a2=2则S3=a2/q+a2+a2q=2/q+2+2q=7(q-2)
s3=a1+a2+a3s3=a1+a1q+a1q^27=1+q+q^2q^2+q-6=0(q-2)(q+3)=0q=2或q=-3(舍去)an=a1q^(n-1)=2^(n-1)
好像无实根啊,题错了?a1+a2+a3=7,\x09a2=7-a1-a3,\x09a22=a12+a32+49+a1a3-7a1-7a3a1xa3=a22=a12+a32+49+a1a3-7a1-7a