数列 an 满足a1=1 3,且前n项的算术平均数 则n的最小值

an=1＋2＋3＋…＋n=[n(n＋1)]/2则：1/(an)=2/[n(n＋1)]=2[(1/n)－1/(n＋1)],所以：M=1/(a1)＋1/(a2)＋1/(a3)＋…＋1/(an)=2[1/1

10Sn=(an)²+5an+610S(n-1)=(a(n-1))²+5a(n-1）+6两式相减,得5a(n-1)+5an=(an)²-(a(n-1))²5=a

2·a(n)=2[Sn-S(n-1)]=(n+1)an-n·a(n-1)∴(n-1)an=n·a(n-1),∴an/[a(n-1)]=n/(n-1),.,a3/a2=3/2,a2/a1=2/1,将上述

a(n+1)=2an-1a(n+1)-1=2(an-1)[a(n+1)-1]/(an-1)=2,为定值.a1-1=3-1=2数列{an}是以2为首项,2为公比的等比数列.an=2×2^(n-1)=2^

a1^3+a2^3+...+an^3=sn^2a1^3+a2^3+...+[a(n+1)]^3=[s(n+1)]^2两式相减得[a(n+1)]^3=[s(n+1)]^2-sn^2[a(n+1)]^3=

因为An=Sn-Sn-1.所以Sn-Sn-1+Sn*Sn-1=0,等式两边同时除以Sn*Sn-1得：1/Sn-1/Sn-1+=1,所以1/Sn为等差数列.因为a1=1/2.所以S1=1/2,1/S1=

设公差为d.7a5=-5a97(a1+4d)=-5(a1+8d)a1=-17代入7(-17+4d)=-5(-17+8d)整理,得4d=12d=3an=a1+(n-1)d=-17+3(n-1)=3n-2

(1)an+2Sn·S(n-1)=0(n≥2),又an=Sn-S(n-1)所以Sn-S(n-1)+2Sn·S(n-1)=0(n≥2)两边同时除以Sn·S(n-1),得1/S(n-1)-1/sn+2=0

an=sn-s(n-1)sn-s(n-1)+2sns(n-1)=01/sn-1/s(n-1)=(s(n-1)-sn)/sns(n-1)=2所以,数列{1/sn}是等差数列,公差是2.首项是1/S1=1

a1=1an=an-1+3n-2an-1=an-2+3(n-1)-2...a2=a1+3*2-2左右分别相加an=a1+3*(n+n-1+...+2)-2*(n-1)an=1+3*(n+2)*(n-1

1.a(n+1)-an=1,为定值,又a1=1,数列{an}是以1为首项,1为公差的等差数列.an=1+n-1=nn=1时,S1+b1=2b1=2b1=1n≥2时,Sn=2-bnS(n-1)=2-b(

1.证：n≥2时,2Sn²=2anSn-an=2[Sn-S(n-1)]Sn-[Sn-S(n-1)]整理,得S(n-1)-Sn=2SnS(n-1)等式两边同除以SnS(n-1)1/Sn-1/S

（1）由已知Sn=n2，得a1=S1=1当n≥2时，an=Sn-Sn-1=n2-（n-1）2=2n-1所以an=2n-1（n∈N*）由已知，b1=a1=1设等比数列{bn}的公比为q，由2b3=b4得

[n]-b[n-1]=(n+1)S[n]/n-nS[n-1]/(n-1)=(通分)=((n²-1)S[n]-n²S[n-1])/n(n-1)∵S[n]-S[n-1]=a[n]∴原式

1、a(n+1)/an=(n+2)/(n+1)a(n+1)/(n+2)=an/(n+1)设cn=an/(n+1)则c(n+1)=a(n+1)/(n+2),且c1=a1/(1+1)=1即c(n+1)=c

因为Sn+Sn-1=3an所以Sn-1+Sn-1+an=3an2Sn-1=2anSn-1=an因为Sn=an+1所以Sn-Sn-1=an+1-anan=an+1-an2an=an+1an+1/an=2

记Tn表示{an}的前n项和a1^3+a2^3+a3^3+...+an^3=(a1+a2+a3+...+an)^2……(1)a1^3+a2^3+a3^3+...+a^3(n-1)=(a1+a2+a3+

(1)6a1=a1^2+3a1+2解得a1=1或2(2)6sn=an^2+3an+26s（n-1）=a（n-1）^2+3a（n-1）+2两式想减得6an=an^2-a(n-1)^2+3an-3a(n-

an=Sn-Sn-1(n>=2)an=1/2a(n-1)-1/2a(n-2)=(1/2)a将a=1代入an不符,则该数列以分段的形式构成an=1(当n=1),an=1/2a(n>=2)

由a(n+1)=(5an-13)/(3an-7)--------1得an=(7a(n+1)-13)/(3a(n+1)-5)又a(n+2)=(5a(n+1)-13)/(3a(n+1)-7)将公式1带入得