在数列an中,sn=3an-1 2

由于:a(n+1)=2an+3则有:[a(n+1)+3]=2(an+3)[a(n+1)+3]/(an+3)=2则:{an+3}为公比为2的等比数列则:an+3=(a1+3)*2^(n-1)=4*2^(

设：(An+1)+p(n+1)+q=4[An+pn+q]解得p=-1,q=0即An+1=4An-3n+1等价于(An+1)-（n+1）=4（An-n）若设Bn=An-n则Bn+1=4Bn则Bn=B1*

由题意Sn1一Sn=1/3SnSn1=4/3Sn所以Sn成等比,首项s1=a1=1,公比为4/3Sn=4/3(n-1)当n≥2时an=Sn-Sn-1=4/3(n-1)一4/3(n-2)所以an=1[n

因为an,Sn,Sn-1/2成等比数列Sn（平方）=an*（Sn-1/2）由an=Sn-S（n-1）Sn（平方）=（Sn-S（n-1））*（Sn-1/2）化简得S（n-1）*Sn=S（n-1）/2-S

S[1]=a[1]=1/2(a[1]+1/a[1]),于是：a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是：a[2]=√2-1,S[2]=√2S[3]=a[3]

∵2√Sn=an+1,∴Sn=(an+1)^2/4∴S(n-1)=(a(n-1)+1)^2/4两式相减,得到an=Sn-S(n-1)=1/4*(an^2-a(n-1)^2)+1/2*(an-a(n-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-Sn-1=4n+1(n>=2),a1=2*1+3=5,满足上式,an通项就是4n+1,即证实等差数列

a(n+1)=3a(n)/[a(n)+3],若a(n+1)=0,则,a(n)=0,...,a(1)=0,与a(1)=1/2矛盾.因此,a(n)不为0.1/a(n+1)=(1/3)[a(n)+3]/a(

an=1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2,a1=1/6所以S1=a1=1/6n>=2时,Sn=a1+a2+...+an=[1/1*2-1/2*3]/2+[1

(1)由2an=Sn*S(n-1),an=Sn-S(n-1)则:2[Sn-S(n-1)]=Sn*S(n-1)2Sn-2S(n-1)=Sn*S(n-1)两边同时除以Sn*S(n-1)2/S(n-1)-2

An=3S(n-1).用原式减去,得A（n+1）-An=3An.A(n+1)=4An.则An为等比数列.

an=Sn-Sn-1=>an=n^2*an-(n-1)^2*an-1an/an-1=(n-1)/n+1)所以an-1/an-2=(n-2)/n)an-2/an-3=(n-3)/n-1)an-3/an-

n≥2时an=Sn-S(n-1)=n²an-(n-1)²a(n-1)∴an/a(n-1)=(n-1)/(n+1)∴a2/a1=1/3a3/a2=2/4a4/a3=3/5……a(n-

an,Sn,Sn-1/2成等比数列an(Sn-1/2)=Sn^2a2(S2-1/2)=S2^2a2(a2+1/2)=(a2+1)^2a2=-2/3a3(S3-1/2)=S3^2a3(a3-1/6)=(

Sn-a1=48,Sn-an=36,Sn-a1-a2-an-1-an=21,∴2Sn-(a1+an)=84Sn-(a1+an)-(a2+an-1)=21∴2Sn-2Sn/n=84Sn-4Sn/n=21

an+1=(1/2)an+1还是an+1=1/(2an)+1如果是前者,用配方法an+1-2=（an-2）/2｛an-2｝是等比数列an-2=(a1-1)/2^(n-1)an=4*2^(-n)+2Sn

Sn=A(n+1)-3n+1Sn+1-Sn=A(n+2)-A(n+1)-3=A(n+1)A(n+2)=2A(n+1)+3令A（n+2）+λ=2[A（n+1）+λ]则A（n+2）=2A（n+1）+λ得λ

因为6Sn=(an+1)(an+2)（1）所以6Sn-1=(an-1+1)(an-1+2)（2）（1）-（2）则an-an-1=3所以an是等差数列因为6Sn=(an+1)(an+2)可知S1＝a1=

Sn=1/3an—2Sn-1=1/3an-1—2Sn—Sn-1=1/3an—1/3an-1an=1/3an—1/3an-1an/an-1=-1/2q=-1/2S1=1/3a1—2a1=1/3a1—22