等差数列an的前n项和均为正数,a1=3

当n=1时,2S1=a1+1/a1,得a1=1当n=2时,2S2=2(1+a2)=a2+1/a2,得a2=√2-1当n=3时,2S3=2(√2+a3)=a3+1/a3,得a3=√3-√2猜想an=√n

2S2=b2(a1+a2)=b1*q*(2a1+d)=32,b3S3=b3(a1+a2+a3)=b1*q²*(3a1+3d)=120,得d=2(都是正数）,q=2.∴an=a1+d(n-1)

a1=3,所以S2=6+d,S3=9+3db1=1,b2=q,b3=q^2所以(6+d)q=64(9+3d)q^2=960相除(9+3d)/(6+d)*q=15q=15(6+d)/(9+3d)代入(6

公差64的等比数列?q=64?→d=-5,这样就很明显了..再问：麻烦看下问题补充再答：公差d,公比q,由已知可得方程（6+d）*q=64，q的d次方=64（{ban}公比64），2个方程解出d=2,

2Sn=an+an^22Sn-1=an-1+an-1^2两式相减：2an=an^2+an-(an-1+an-1^2)an^2-an-(an-1+an-1^2)=0(an-(an-1+1))(an+an

a3=1+2db3=3q²所以1+2d+3q²=17T3=b1+b2+b3=3+3q+3q²S3=a1+(a1+d)+(a1+2d)=3+3d所以3+3q+3q²

因为2Sn=an^2+n-4,所以2S（n-1）=a（n-1）²+n-1-4.两式相减2an=an^2-a（n-1）²+1,a（n-1）²=an^2-2an+1=（an-

an,sn,an^2成等差数列,则2sn=an^2+an那么2s(n-1)=a(n-1)^2+a(n-1)俩式相减：2sn-2s(n-1)=an^2+an-a(n-1)^2-a(n-1)而an=sn-

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)=

sn=an(an+1)/2s(n-1)=a(n-1)(a(n-1)+1)/2两式相减an=an(an+1)/2-a(n-1)(a(n-1)+1)/2an^2-an-a^2(n-1)-a(n-1)=0(

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

sn=(1/8)(an+2)²S(n-1)=(1/8)[a(n-1)+2]²an=Sn-S(n-1)=(1/8){(an+2)²-[a(n-1)+2]²}=(1

4Sn=2an+an^24S(n-1)=2a(n-1)+a(n-1)^2相减得4an=2an-2an(n-1)+[an+a(n-1)][an-a(n-1)]2[an+a(n-1)]=[an+a(n-1

∵{log2an}是公差为-1的等差数列∴log2an=log2a1-n+1∴an=2log2a1−n+1=a1•2−n+1∴S6=a1（1+12+…+132）=a1•1−1261−12=38，∴a1

（1）2Sn=an+an²①2S（n-1）=a（n-1）+a（n-1）²②②-①2Sn-2S（n-1）=an+an²-a（n-1）-a（n-1）²2an=an+

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)=