已知数列an为等差数列,Sn为其前n项和,且2Sn=an 2n

1、Sn=(a1+an)n/2所以nan/Sn=2an/(a1+an)=2[a1+(n-1)d]/[2a1+(n-1)d]上下除以(n-1)=2[a1/(n-1)+d]/[2a1/(n-1)+d]n-

1、Sn=(a1+an)n/2所以nan/Sn=2an/(a1+an)=2[a1+(n-1)d]/[2a1+(n-1)d]上下除以(n-1)=2[a1/(n-1)+d]/[2a1/(n-1)+d]n-

1+Sn=2anSn=2an-1n>=2则S(n-1)=2a(n-1)-1相减an=2an-2a(n-1)an=2a(n-1)所以n>=2时是等比q=2a1=S1所以1+a1=2a1a1=1所以an=

因为Sn-Sn-1=n^2-3n-{(n-1)^2-3(n-1)}=2n-4.又由an=Sn-Sn-1,所以an=2n-4,最后还要验证一下,当n=1时,S1=a1,符合题意.d=an-an-1=2易

∵Sn-Sn-1=-anSn-Sn-1=-an/2∴d=1/Sn-1/Sn-1=（Sn-Sn-1）/SnSn-1=21/S1=1/a1=2∴{1/Sn}为首项=2,公差=2的等差数列

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

a1=1,a2=3/2,a3=7/4an=2-(1/2)^(n-1)Tn=n/(2^(n-1))+(1/2)^(n-2)+n^2+n-4

由Sn=Sn-1/2Sn-1+1,两边同时取倒数可得1/Sn=(2Sn-1+1)/Sn-11/Sn=2+1/Sn-1即1/Sn-1/Sn-1=2故{1/Sn}是首项为1/2,公差为2的等差数列1/Sn

看图

∵1,an,Sn为等差数列∴2a1=1+S1=1+a12a2=1+S2=1+a1+a2∴a1=1a2=2由2an=1+Sn2a(n-1)=1+S(n-1)得2an-2a(n-1)=Sn-S(n-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)=

a3=a1+2d=6S3=a1+a2+a3=3a1+3d=12解得a1=2,d=2,故an=2n所以Sn=n(n+1)所以1/S1+1/S2+……+1/Sn=1/(1*2)+1/(2*3)+1/(3*

S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+

Sn=n(an+1)/2S(n+1)=(n+1)[a(n+1)+1]/2用下式减上式a(n+1)=[(n+1)a(n+1)-nan+1]/2即2a(n+1)=[(n+1)a(n+1)-nan+1]即(

当n≥2时,可以化为Sn-S(n-1)=-2Sn×S(n-1),两边同除以Sn×S(n-1),得1／Sn-1／S(n-1)=2所以｛1/Sn｝是以2为首项,2为公差的等差数列即1/Sn=2nSn=1/

当n=1时,a1=S1=1当n≥2时,an=Sn-S(n-1)=3n²-2n-3(n-1)²+2(n-1)=6n-5∵当n=1时,满足an=6n-5又∵an-a(n-1)=6n-5

数列{an+Sn}是公差为2的等差数列∴an+Sn=a1+s1+2(n-1)=1+1+2n-2=2n∵当n=2时,a2+a2+a1=4∴a2=3/2(2)当n>=2时由an+Sn=2n得a(n-1)+

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)=