在数列an中,Sn=4n^2-n

Sn=(an+1/an)s1=a1=(a1+1/a1)/2==>a1=1/a1==>a1=1(由于an>0所以a1=-1不合题意)s2=a1+a2=(a2+1/a2)/2==>2a1+a2=1/a2将

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*

a(n+1)=a(n)+2说明这是一个等差数列首项a(1)=-11,公差为2a(n)=a(1)+(n-1)×2=-11+2(n-1)=2n-13所以Sn=[a(1)+a(n)]×n/2=(n-12)n

（1）当n=1时,S2-4S1=0,又因为a1=1,所以S1=1,既S2=4S1=4a1=4S2=a1+a2=4既：a2=3当n=1时,又2a（n+1）为bn与b（n+1）的等比中项所以bn*b(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

题目是不是消失了an=2S(n-1)an=Sn-S(n-1)Sn-S(n-1)=2S(n-1)Sn=3S(n-1)则：{Sn}是等比数列S1=a1=1公比q=3Sn=3^(n-1)an=2S(n-1)

(Sn)²=[Sn-S(n-1)](Sn-1/2)(Sn)²=(Sn)²-Sn/2-SnS(n-1)+S(n-1)/2Sn+2SnS(n-1)-S(n-1)=0S(n-1

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

（1）证明：∵Sn-2an=2n，①∴Sn+1-2an+1=2（n+1）．②②-①，得：an+1-2an+1+2an=2，∴an+1=2an-2，∴an+1-2an-2=(2an-2)-2an-2=2

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

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)=(

将a[n+1]=S[n+1]-S[n]代人得到：S[n]=4(S[n+1]-S[n])+14S[n+1]=5S[n]-14(S[n+1]-1)=5(S[n]-1)(S[n+1]-1)/(S[n]-1)

Sn=n^2+4nS(n-1)=(n-1)^2+4(n-1)=n^2+2n-3An=S(n)-S(n-1)=2n+3

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

Sn=3n^2-2n当n>=2,sn-1=3(n-1)^2-2(n-1)an=sn-sn-1=3n^2-2n-3(n-1)^2+2(n-1)=6n-5(n>=2)当n=1,s1=a1=1满足上个式子所

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