已知数列an Sn是其前n项的和 a1=2 Sn 1=3Sn n^2 2

∵数列{an}的通项公式是an=2n−12n，∴an=1-12n，∴Sn=（1-12）+（1-14）+（1-18）+…+（1-12n）=n-（12+14+18+…+12n）=n-12[1−(12)n]

因为S(n+1)-S(n)=A(n+1),根据题意有:2S(n+1)^2=2A(n+1)S(n+1)-A(n+1),将上式代入此式得:2S(n+1)^2=2[S(n+1)-S(n)]S(n+1)-S(

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

已知：数列{an}中,a1=1,Sn为数列{an}的前n项和,且2an/(anSn-Sn²)=1,(n≥2)；(1).证明数列{1/Sn}是等差数列；(2).求数列{an}的通项公式.(1)

an=(2^n-1)/2^n=1-1/2^n令b1=1/2公比q=1/2设bn=1/2^1+1/2^2+1/2^3...+1/2^n则bn=b1(1-q^n)/(1-q)=.=1-1/2^nSn=a1

Sn=4(4^n-1)/(4-1)-2(2^n-1)/(2-1)=[4^(n+1)-4)/3-[2^(n+1)-2]=[4^(n+1)-4-3*2^(n+1)+6]/3=[2^(n+1)*2^(n+1

(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6

1.证：n≥2时,2Sn²=2anSn-an=2[Sn-S(n-1)]Sn-[Sn-S(n-1)]整理,得S(n-1)-Sn=2SnS(n-1)等式两边同除以SnS(n-1)1/Sn-1/S

2Sn的平方=2anSn-an2Sn(Sn-an)=-an2SnS(n-1)=S(n-1)-Sn1/Sn-1/Sn-1=2｛1/Sn｝等差数列公差2首项1/21/Sn=1/2+2(n-1)=[4n-3

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*

a3=-13(a1+a9)*9/2=-45a1+a9=-10所以a1+2d=-13,2a1+8d=-10所以a1+2d=-13,a1+4d=-5解得d=4a1=-21an=-21+4(n-1)=4n-

1.证：Sn=(3an-n)/2Sn-1=[3a(n-1)-(n-1)]/2an=Sn-Sn-1=[3an-3a(n-1)-1]/2an=3a(n-1)+1an+1/2=3a(n-1)+3/2=3[a

n=1时,(s1-1)^2=s1*s1即-2s1+1=0解得s1=1/2n=2时,(s2-1)^2=(s2-s1)*s2解得：s2=2/3n=3时,(s3-1)^2=(s3-s2)*s3解得：s3=3

2(S_n)^2=2a_nS_n-a_n=>2S_n(S_n-a_n)=-a_n=>2S_n*S_{n-1}=-a_n2S_n*S_{n-1}=-(S_n-S_{n-1})2=-1/S_{n-1}+1

由题意知：2an/[anSn-(Sn)²]=1（n>1）则：(Sn)²-anSn+2an=0（n>1）又因为：an=Sn-S(n-1)（n>1）所以：(Sn)²-[Sn-

因为Sn=3n^2+5nS（n-1）=3（n-1）^2+5(n-1)两式相减所以an=6n-3+5=6n+2所以an=8+6（n-1）,所以an是以8为第一项,公差为6的等差数列.

∵数列{an}的通项公式an=2n+1，∴Sn=n(3+2n+1)2=n2+2n，∴Snn=n+2，∴数列{Snn}的前10项的和为10(3+12)2=75．故答案为：75．

证明：（1）∵an2-2anSn+1=0，an=Sn-Sn-1（n≥2）∴（Sn-Sn-1）2-2（Sn-Sn-1）Sn+1=0⇒Sn2-Sn-12=1故{Sn2}成等差数列．（2）∵a12-2a12

证::n=1,a1=s1=4n>1an=Sn-Sn-1Sn=n^2+3nSn-1=(n-1)^2+3(n-1)an=2n+2经验证n=1满足通项n>1an-an-1=2,由等差数列定义可知,数列{an

（1）3an=2Sn+n...①3an+1=2Sn+1+n+1...②②-①得：3an+1-3an=2an+1+1即an+1=3an+1==＞an+1+1/2=3(an+1/2)an+1+1/2/an