Sn=n² 2n公比大于1
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Sn=a1*(1-q^n)/(1-q),Tn=a1*(1-q^2)/(1-q)q不等于1时,lim(Sn)/Tn=lim(1-q^n)/(1-q^2n)q1,lim(Sn)/Tn=lim1/q^n=0
an=a1q^(n-1)=(1/2)^(n-1)Sn=1+1/2+(1/2)^2+……+(1/2)^(n-1)Sn/2=1/2+(1/2)^2+……+(1/2)^(n-1)+(1/2)^n二式的两边相
设等比数列{an}的公比为q,前n项和为Sn,且Sn+1,Sn,Sn+2成等差数列,则2Sn=Sn+1+Sn+2.若q=1,则Sn=na1,式子显然不成立.若q≠1,则有2a1(1−qn)1−q=a1
(1)∵{An}为等比数列,则有An+1=An·q,又∵Sn+1,Sn,Sn+2成等差数列,∴Sn+1+Sn+2=2Sn∴Sn+An+Sn+An+An·q=2Sn∴可得2+q=0所以q=-2(2)这里
a(n)=aq^(n-1),n=1,2,...若q=1.则s(n)=na,n=1,2,...s(n+1)+s(n+2)-2s(n)=(n+1)a+(n+2)a-2na=3a不等于0,矛盾.因此,q不为
因为Sn+1,Sn,Sn+2成等差数列S(n+1)+S(n+2)=2*S(n)(q^(n+1)-1)*a1/(q-1)+(q^(n+2)-1)*a1/(q-1)=2*(q^(n)-1)*a1/(q-1
s4/a4=[a1(1-q^4)/(1-q)]/a1q^3=[(1-q^4)/(1-q)]/q^3=[(1-q)(1+q)(1+q^2)]/(1-q)]/q^3=(1+q)(1+q^2)/q^3=(1
s4=a1(1-q^4)/(1-q)a4=a1q^3s4/a4=(1-q^4)/q^3(1-q)=(1-1/16)/(1/16)=15再问:
Sn=2(1-3^n)/(1-3)=3^n-1S(n+1)=3*3^n-1S(n+1)/Sn=(3*3^n-1)/(3^n-1)=(3*3^n-3+2)/(3^n-1)=3+2/(3^n-1)(3n+
证:a1=2q=3Sn=2(3^n-1)/(3-1)=3^n-1Sn+1=2[3^(n+1)-1]/(3-1)=3^(n+1)-1Sn+1/Sn=[3^(n+1)-1]/(3^n-1)=[3^(n+1
【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=
依据题意,有2*3a2=a1+3+a3+4=7+a1+a3=7+a1+a2+a3-a2=7+7-a2=14-a2.2*3a2=14-a26a2=14-a27a2=14.a2=2.s3=a1+a2+a3
an=-Sn.S(n-1)Sn-S(n-1)=-Sn.S(n-1)1/Sn-1/S(n-1)=11/Sn-1/S1=n-11/Sn=nSn=1/n
设首项为a1,公比为r,当r=1时,Sn=n(a1),此时Sn/S(n+1)的极限为1r≠1时,Sn=a1(1-r^n)/(1-r),Sn/S(n+1)=(1-r^n)/(1-r^(n+1)),极限为
an=sn-s(n-1)代入得Sn=2S(n-1)+2^n,即Sn/2^n=S(n-1)/2^(n-1)+1所以Sn=(n+1/2)*2^n,所以an=Sn-S(n-1)=n*2^n+2^(n-1).
n=1/(n(n+1))+2^(2n-1)=1/n-1/(n+1)+2^2n/2=1/n-1(n+1)+1/2*4^nE此数列依次从1到n,消掉得=1-1/(n+1)E此数列是等比数列得1/2*=2*
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)