a1=1,an=2Sn

如图

a1=1,Sn=a1+a2+a3+……+an,an=2S(n-1)Sn=a1+a2+...+a(n-1)+an=S(n-1)+2S(n-1)=3S(n-1)Sn=3^(n-1)Sn-S(n-1)=an

an+2Sn*Sn-1=0其中an=Sn-Sn-1代入上式：Sn-Sn-1+2Sn*Sn-1=0a1＝1/2,故Sn和Sn-1≠0,上式两边同除以Sn*Sn-1得：1/Sn-1-1/Sn+2＝0即：1

解题思路：将an用Sn－S(n-1)表示，整理得到Sn与S(n-1)的关系，归结为等差数列的定义形式解题过程：数列{an}的首项an=1，前n项和sn之间满足，求证{1/sn}成等差数列；并求Sn的表

an=-Sn.S(n-1)Sn-S(n-1)=-Sn.S(n-1)1/Sn-1/S(n-1)=11/Sn-1/S1=n-11/Sn=nSn=1/n

由题意可得an=2Sn^2/(2Sn-1)又由于an=Sn-S(n-1)即Sn-S(n-1)=2Sn^2/(2Sn-1)化简得Sn+2SnS(n-1)-S(n-1)=0两边同除SnS(n-1)得1/S

由题意得：2S(n+1)=4Sn+a1,则2Sn=4S(n-1)+a1解得：a(n+1)=2an,则{an}为等比数列,公比q=2所以,an=a1q^(n-1)=2^n同样：2S(n+1)=4Sn+a

已知a_(n+1)=S_n得a_n=S_(n-1)（n＞1）两式相减a_(n+1)-a_n=S_n-S_(n-1)=a_n（n＞1）得a_(n+1)=2a_n（n＞1）因为a_2=S_1=a_1=-2

我会我会Sn+1=Sn-2nSn+1Sn两边同除以Sn+1*Sn得1/Sn+1-1/Sn=2n以此类推1/Sn-1/Sn-1=2(n-1)1/Sn-1-1/Sn-2=2(n-2)...1/S2-1/S

由题意,S(n)-S(n-1)=2a(n+1)-2a(n),即a(n)=2a(n+1)-2a(n),于是a(n+1)=a(n)*3/2,即a(n)是公比是q=3/2的等比数列,且首项是a(1)=1,所

n>=2时：∵an=2Sn^2/[（2Sn）-1]∴Sn-(Sn-1)=2Sn^2/[（2Sn）-1]两边同时乘以（2Sn）-1并化简得2Sn(Sn-1)+Sn-(Sn-1)=0两边同时除以Sn(Sn

n≥2时,an=Sn-S(n-1)=2Sn²/(2Sn-1)[Sn-S(n-1)](2Sn-1)=2Sn²-Sn-2SnS(n-1)+S(n-1)=0S(n-1)-Sn=2SnS(

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

Sn-1=(n-1)(n-1)an-1Sn-Sn-1=an=nnan-(n-1)(n-1)an-1(nn-1)an=(n-1)(n-1)an-1an=(n-1)/(n+1)*(n-2)/(n-1)*…

a[n+1]=a[n]/（a[n]+2）是不是这样子?那么两边同时取倒数.1/a[n+1]=[an+2]/an=1+2/an1/a[n+1]+1==2+2/an=2{1/an+1}所以形如1/an+1

因为n,an,Sn成等差数列所以2an=Sn+n又因为an=Sn-Sn-1所以Sn+n=2Sn-1+2n左右两边同时加2Sn+n+2=2Sn-1+2n+2右边再变化Sn+n+2=2Sn-1+2n+2-

2(Sn+1)(Sn)/(Sn-Sn+1)=1上下除以(Sn+1)(Sn)得到2/(1/Sn+1-1/Sn)=11/(Sn+1)-1/Sn=2因此1/Sn+1为等差数列,1/S1=1/a1=1/21/

∵s[n]=n^2a[n]∴s[n+1]=(n+1)^2a[n+1]将上述两式相减,得：a[n+1]=(n+1)^2a[n+1]-n^2a[n](n^2+2n)a[n+1]=n^2a[n]即：a[n+

/>n≥2时,Sn=n²×anS(n-1)=(n-1)²×a(n-1)an=Sn-S(n-1)=n²×an-(n-1)²×a(n-1)(n²-1)an

证明如下：