已知an的各项均为正数,前n项和为Sn且满足2Sn=an² n-4

（1）∵2Sn=an2+n-4（n∈N*）．∴2Sn+1=an+12+n+1-4．两式相减得2Sn+1-2Sn=an+12+n+1-4-（an2+n-4），即2an+1=an+12-an2+1，则an

设公比为q,则q＞0,a2=a1q,a3=a1q²由a1=2,S3=a1+a2+a3=14,得q=2∴an=2^n(n为正整数)证明：bn=n/an=n/2^n(n为正整数),前n项和为Tn

4Sn=(an+1)^24Sn-1=(an-1+1)^2n-1为下标则4an=4Sn-4Sn-1=(an+1)^2-(an-1+1)^2化简得(an-1)^2=(an-1+1)^2则an-1=正负（a

证明：∵Sn=an(an+1)2∴S1=a1(1+a1)2∴a1=1…（1分）由2Sn=a2n+an2Sn-1=a2n-1+an-1⇒2an=2(Sn-Sn-1)=a2n-a2n-1+an-an-1…

1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=

n=1时,2a1=2S1=a1^2+1-4a1^2-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=an^2+n-4-a

4a(1)=[a(1)+1]^2a(1)=14a(n+1)=[a(n+1)+1]^2-[a(n)+1]^2[a(n)+1]^2=[a(n+1)-1]^2若a(n+1)>1a(n+1)=a(n)+2a(

6Sn=an^2+3an+26S(n-1)=a(n-1)^2+3a(n-1)+26Sn-6S(n-1)=6an=an^2+3an+2-a(n-1)^2-3a(n-1)-26an=an^2+3an-a(

1)6Sn=An^2+3An+2因为S1=A1所以6A1=A1^2+3A1+2A1^2-3A1+2=0(A1-1)(A1-2)=0因为A1=S1>1所以A1=2因为An=Sn-S(n-1)注S(n-1

当n=1时,S1=a1=1/2(a1^2+a1),解得a1=1当n＞1时,an=Sn-S(n-1)=1/2(an^2+an)-1/2[a(n-1)^2+a(n-1)],整理得[an+a(n-1)][a

6Sn=An^2+3An+26S(n-1)=[A(n-1)]^2+3A(n-1)+26Sn-6S(n-1)=6An=An^2+3An+2-{[A(n-1)]^2+3A(n-1)+2}An-A(n-1)

（1）a1＝(a1+1)24，解得a1=1，当n≥2时，由an=Sn-Sn-1=(an+1)2−(an−1+1)24，得（an-an-1-2）（an+an-1）=0，又an＞0，所以an-an-1=2

（1）当n=1时，a1＝s1＝14a21+12a1−34，解出a1=3，又4Sn=an2+2an-3①当n≥2时4sn-1=an-12+2an-1-3②①-②4an=an2-an-12+2（an-an

n>=2时,S[n]=1/4*(a[n]+1)^2;S[n-1]=1/4*(a[n-1]+1)^2两式相减得到a[n]=1/4*(a[n]^2+2a[n]-a[n-1]^2-2a[n-1])化简得到a

4a(1)=[a(1)+1]^2a(1)=14a(n+1)=[a(n+1)+1]^2-[a(n)+1]^2[a(n)+1]^2=[a(n+1)-1]^2若a(n+1)>1a(n+1)=a(n)+2a(

根号Sn的通项公式是nSn=n^2an=Sn-Sn-1=n^2-(n-1)^2=2n-1

Sn、an、1成等差,则2an=Sn＋1(n=1时,得a1=1),当n≥2时,有2a(n－1)=S(n－1)＋1,则2an－2a(n－1)=an,即an/[a(n－1)]=2=常数,所以{an}是等比

由题意知2an=Sn+1/2,an＞0,当n=1时,2a1=a1+1/2,解得a1=1/2,当n≥2时,Sn=2an-1/2,S(n-1)=2a(n-1)-1/2,两式相减得an=Sn-S(n-1)=

（1）由Sn，an，12成等差数列，可得2an＝Sn+12，∴a1＝12，a2=1（2）由2an＝Sn+12可得，2Sn=4an-1（n≥1），∴2Sn-1=4an-1-1（n≥2）∴两式相减得2an

由题意2an=Sn+1/2Sn=2an-1/2n=1时,S1=a1a1=2a1-1/2a1=1/2S(n+1)-Sn=a(n+1)2a(n+1)-1/2-[2an-1/2]=a(n+1)a(n+1)=