an 为等比数列前n项和为Sn,S3,S9,S6成等差数列

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

n,an,Sn成等差数列,所以n+Sn=2an,即Sn=2an-n,an+1=Sn+1-Sn=2an+1-n-1-2an+n=2an+1-2an-1化简就是an+1=2an+1an+1+1=2an+2

（Ⅰ）∵等比数列{an}的前n项和为Sn，S1，S3，S2成等差数列，∴2（a1+a1q+a1q2）=a1+a1+a1q，解得q=-12或q=0（舍）．∴q=-12．（Ⅱ）∵a1-a3=3，q=-12

an=a1+（n-1）dS1=a1S2=S1q=a1+a2=2a1+d.1S4=S1q^2=a1+a2+a3+a4=4a1+6d.21式、2式两边都除以a1,得q=2+d/a1,q^2=4+6*(d/

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-

S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S1,S3,S2成等差数列即s3-s1=s2-s31+q+q^2-1=1+q-(1+q+q^2)q^2+q=-q^2q=0或-1/2如果a1-

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

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

Sn=n-5an-85则an=Sn-S(n-1)=n-5an-85-(n-1)+5a(n-1)+85=1-5an+5a(n-1)即6an=5a(n-1)+16an-6=5a(n-1)+1-66(an-

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）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

这个直接用a5=s5-s4=(32+r）-（16+r）=16

∵Sn=kq^n-k∴S(n+1)=kq^(n+1)-k∴a(n+1)=S(n+1)-Sn=[kq^(n+1)-k]-(kq^n-k)=k[q^(n+1)-q^n]=k[(q-1)q^na(n+1)/

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

已知等比数列an,首项为81,数列bn满足bn=log3an,其前n项和sn(1)证明：bn-b(n-1)=log(3)an-log(3)an-1=log（3）an/a(n-1)=log(3)q∵b1