设等比数列an前n项和为sn,若s2＝3

S4=a1+a2+a3+a4=a2/q+a2+a2*q+a2*q^2S4/a2=1/q+1+q+q^2=7.5

an=Sn+1-Sna1=b1=S1a2=S2-S1b2=b1/(a2-a1)因为bn是等比数列,所以b2就知道了然后cn的通项公式就知道后面的应该没啥大问题只授剑意,不授剑招

∵Sn=n-an,∴a(n+1)=S(n+1)-S(n)=(n+1)-a(n+1)-n+a(n)=1+a(n)-a(n+1);∴2a(n+1)=1+a(n)；∴2a(n+1)-2=1+a(n)-2,即

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

若q=1Sn=nA1Un=(A1+nA1)×n/2=(n+1)nA1/2若q≠1Sn=A1×(1-q^n)/(1-q)=A1/(1-q)-A1/(1-q)×q^nUn=nA1/(1-q)-A1/(1-

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

S1,S2...,Sn...的首项S1=a1,设公比为q,an≠0,所以q≠1.则Sn=a1q^(n-1),(n=1,2,...),S(n-1)=a1q^(n-2),(n=2,3,...),相减,an

不一定,当S1,S2,S3.Sn都相等时,a2,a3.an为0数列,不成等比．当S1,S2,S3.Sn公比不为1时,an=sn-s(n-1)不为0,则有a(n+1)/an=[s(n+1)-s(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

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

a(n+1)=2S(n-1)(1)a(n)=2S(n-2)(2)a(n+1)-an=2a(n-1)a(n+2)-a(n+1)-2an=0Theauxilaryequationx^2-x-2=0(x-2

设{an}的公比为q，由S4=1，S8=17知q≠1，∴得a1(q4−1)q−1＝1①a1(q8−1)q−1＝17②由①和②式整理得q8−1q4−1＝17解得q4=16所以q=2或q=-2将q=2代入

Sn=a1(1-q^n)/(1-q)Sn+1=a1[1-q^(n+1)]/(1-q)Sn+2=a1[1-q^(n+2)]/(1-q)2Sn+2=Sn+Sn+1a1[1-q^(n+1)]/(1-q)+a

证明：A（n+1）=Sn+3n+1,则An=S（n-1）+3n-2两式想减得A（n+1）-An=Sn+3n+1-（S（n-1）+3n-2）=An+3即A（n+1）+3=2（An+3）即（A（n+1）+

不成等比数列∵s1,s2,.sn成等比数列则S1,S2,S3必有S1*S3=S2^2即a1*(a1+a2+a3)=(a1+a2)^2化简得a1a3=a2^2+a1a2①若a1,a2..成等比数列成立必

Sn=4An-3S(n-1)=4A(n-1)-3Sn-S(n-1)=An=4An-3-[4A(n-1)-3]=4an-3-4A(n-1)+3=4An-4A(n-1)3An=4A(n-1)An/A(n-

an=Sn-S(n-1)=2(an-3)-2[a(n-1)]-3=2an-2a(n-1)]an=2a(n-1)所以an是等比数列q=1S1=a1所以a1=2(a1-3)a1=6所以an=6*2^(n-

Sn=3a(n+1)+m与S(n-1)=3an+m两式相减：Sn-S(n-1)=an=3a(n+1)-3an.a(n+1)/an=4/3,所以q=4/3.

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