等比数列{an},前n项和Sn=3*2^n m,则其公比是

Sn=na1,q=1a1(1-q^n)/(1-q)=(a1-anq)/(1-q),q≠1

由Sn=3^(n+1)+r可知公比q=3取n=1得a1=9+r取n=2得a1+a2=4a1=27+r解得a1=6,r=-3

设公比为q，当q=-1时，等比数列{an}的各项是a，-a，a，-a，a，-a…的形式，a≠0．又已知Sn是实数等比数列{an}前n项和，故当n为偶数时，Sn=0，当n为奇数时，Sn=a，故选D．

等比数列{an}中,前n项和为sn,已知S1,S3,S2成等差数列,求{an}的公比Q.已知a1-a3=3,求sn?S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S1,S3,S2成等差数列

设等比数列{a[n]}的公比为q则S[n]=a[1](1-qⁿ)/(1-q)=2(1-qⁿ)/(1-q)则S[n]+1=2(1-qⁿ)/(1-q)+1S[1]+1=

an+sn=-2n-1,当n＝1时,a1+s1=-3,则a1=-3/2.由已知得：sn=-2n-1-an当n大于或等于2时,则an=sn-s(n-1）=-2n-1-an-[-2(n-1)-1-a(n-

n=1时,a1=1+3a1.即a1=-1/2.n>1时,an=Sn-Sn-1=1+3an-（1+3a(n-1））=3an-3a(n-1),即an=3/2a(n-1),即an=-1/2*(3/2)^(n

Sn=n-5an-85S1=1-5a1-85即a1=1-5a1-85解得a1=-14an=Sn-S(n-1)=n-5an-85-[(n-1)-5a(n-1)-85]=-5an+5a(n-1)+16an

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

Sn=2an-3n+5S(n-1)=2a(n-1)-3(n-1)+5相减an=2a(n-1)+3an+3=2a(n-1)+6an+3=2[2a(n-1)+3]

求出首项a1和公比q代入公式就可以了当q≠1时an=a1q^(n-1)sn=a1(1-q^n)/(1-q)当q=1时an=a1sn=na1

为了避免混淆,我把下角标放在内.首先从数列本身的基本意义出发a=S-S其次,从已知a=S(n+2)/n出发a=S*(n+1)/(n-1)因此S-S=S*(n+1)/(n-1)移项整理S=S

1、A(n+1)=(n+2)sn/n=S(n+1)-Sn即nS(n+1)-nSn=(n+2)SnnS(n+1)=(n+2)Sn+nSnnS(n+1)=(2n+2)SnS(n+1)/(n+1)=2Sn/

1）设an=a1*q^(n-1),则有Sn=a1*(1-q^n)/(1-q),[Sn*Sn+2-（Sn+1）^2]=a1^2*{(1-q^n)*[1-q^(n+2)]-[1-q^(n+1)]^2}/(

数列{an}前N项和Sn3Sn=(an-1),(1)当n>=2,有:3Sn-1=[a(n-1)-1],(2)(1)-(2),3an=an-an-1an/an-1=-1/2,(n>=2)当n=1,3S1

n=b1.q^(n-1)bn=an-3nan=bn+3n=b1.q^(n-1)+3nSn=a1+a2+...+an=b1(q^n-1)/(q-1)+3n(n+1)/2

a1=S1=3+a,a2=s2-s1=9+a-(3+a)=6,a3=S3-S2=27+a-(9+a)=18,因为a1·a3=a2²,即18(3+a)=36,解得a=-1再问：求通项公式再答：

已知Sn=2An-1取n=1得:S1=2A1-1又因为S1=A1,解上述方程可得:A1=1Sn=2An-1S(n-1)=2A(n-1)-1注:"n-1"为下标上下两式相减得:Sn-S(n-1)=2An

（1）令n=1,得a1=-1.Sn=2an+n,S(n+1)=2a(n+1)+n+1.两式相减,得a(n+1)=2a(n+1)-2an+1.整理得a(n+1)-1=2(an-1),a1-1=-2.综上

a1=s1=5-aan=sn-s(n-1)=5^n-a-(5^(n-1)-a)=5^n-5^(n-1)=4*5^(n-1)当n=1时,an=4即5-a=4a=1