等比数列前n项和为sn一直对任意的

n=1时s1=a1=b+rn=2时S2=a1+a2=b²+rn=3时S3=a1+a2+a3=b³+ra1=b+rs2-s1=a2=b²-bs3-s2=a3=b³

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

1.(n,Sn)代入y=b^x+rSn=b^n+rn>=2时An=Sn-S(n-1)=b^n+r-b^(n-1)-r=(b-1)×b^(n-1)要使{An}为等比数列,A1也需满足上式A1=S1=b+

解（1）由题意Sn=b^nr①Sn1=b^n1r②做差得An1=b^n1-b^n公比为b比较A1和A2可得r=-1（2）b=2时An=2^(n-1)得Bn=(n1)/2^(n1)利用错位相消法Tn=(

Sn=b^n+rSn-1=b^n-1+r两式相减an=b^(n-1)*(b-1)故a1=b-1故S(1)=b-1将点(1,b-1)代入得r=-1整理知b(n)=(n+1)/2^(n+1)T(n)=b(

1.(n,Sn)代入y=b^x+rSn=b^n+rn>=2时An=Sn-S(n-1)=b^n+r-b^(n-1)-r=(b-1)×b^(n-1)要使{An}为等比数列,A1也需满足上式A1=S1=b+

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

解题思路：利用函数的性质解题过程：

∵a(n+1)=（n+2)Sn/n且a(n+1)=S(n+1)-Sn∴S(n+1)-Sn=(n+2)*Sn/n∴S(n+1)=[(n+2)/n+1]Sn=(2n+2)/n*Sn∴S(n+1)/(n+1

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

Sn=a1(1-q^n)/(1-q)=a1/(1-q)-a1/(1-q)*(q^n)设b=a1/(1-q),有Sn=b-b*q^n估计你已经会了S1+.+Sn=nb-b(q(1-q^n)/(1-q))

显然k=-1,等比数列前n项和中指数式的系数和常数项互为相反数

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

an^2/a(n+1)=an^2/(an×q）=an/qTn=a1/q+……+an/q=Sn/q当Sn

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