已知数列an的首项为1,其前n项的和为sn,且满足an=2sn² 2sn-1

Sn=n^2*an,a1=1/2当n≥2时有S(n-1)=(n-1)^2*a(n-1)所以an=Sn-S(n-1)=n^2*an-(n-1)^2*a(n-1)即(n^2-1)an=(n-1)^2*a(

因为An+1=2SnAn=2S(n-1)所以A(n+1)-An=2AnA(n+1)/An=3是公比为3,首项a1=1的等比数列,An=A1*q^(n-1)即An=3^(n-1)

因an=log2[(n+2)/(n+1)]=log2(n+2)-log2(n+1),n应该从1开始.所以Sn=log2(3)-log2(2)+log2(4)-log2(3)+...+log2(n+2)

∵Sn=n²+n+1,∴a1=S1=1+1+1=3,当n≥2时,an=Sn-Sn-1(n-1是下标)=（n²+n+1）-[(n-1)²+(n-1)+1]=2n．当n=1时

Sn=5n^2-3n-1S(n-1)=5(n-1)^2-3(n-1)-1=5n^2-13n+7a(n)=S(n)-S(n-1)=10n-8n≥2a(1)=S(1)=5-3-1=1故a(n)=10n-8

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

/>错位相减求和Sn=1/2^1+3/2^2+5/2^3+.+(2n-3)/2^(n-1)+(2n-1)/2^n①‘①×1/2(1/2)Sn=1/2^2+3/2^3+.+(2n-3)/2^n+(2n-

Sn-S(n-1)-2^n=S(n-1)Sn/2^n-S(n-1)/2^(n-1)=1S1=1soSn/2^n=nSn=n*2^nan=Sn-S(n-1)=n*2^n-(n-1)2^(n-1)an/2

an=(2^n-1)/2^n=1-(1/2)^nSn=n-1/2(1-(1/2)^n)/(1-1/2)=n-1+(1/2)^n=321/64解得n=6

（1）2Sn=an^2+an2Sn-1=a(n-1)^2+a(n-1)2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)an^2-a(n-1)^2=an+a(n-1)[an+a

因为：An+1=2Sn,则A(n-1)+1=2S(n-1)那么：2Sn-2S(n-1)=（An+1）-（A(n-1)+1）（n>=2）又因为：2Sn-2S(n-1)=2An（n>=2）所以：2An=（

(2n+1)^2-(2n-1)^2=4n^2+4n+1-(4n^2-4n+1)=8nAn=[(2n+1)^2-(2n-1)^2]/[(2n-1)^2(2n+1)^2]=(2n+1)^2/[(2n-1)

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

an=3n-17再问：怎知an=3n-17

1.数列的第n项：a(n)=S(n)-S(n-1)=2a(n)-2a(n-1)移项得a(n)=2*a(n-1)所以n≥2时数列{a(n)}为公比q=2的等比数列;a(2)=S(2)-S(1)=2a(2

解题思路：方法：数列通项的求法：已知sn,求an。求和：错位相减法。解题过程：

∵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．

an=n(2^n-1)an=n*2^n-na1=1*2^1-1a2=2*2^2-2a3=3*3^3-3.an=n*2^n-nSn=a1+a2+a3+.+an=1*2^1-1+2*2^2-2+3*3^3

an=log2(n+1)-log2(n+2)Sn=log2(2)-log2(3)+log2(3)-log2(4)+.+log2(n)-log2(n+1)+log2(n+1)-log2(n+2)=log