已知sn是等比差不数列an的前n项和s3 s6=1 3求s6 s12=?

n≥2时,Sn=4a(n－1)＋2,与S(n＋1)=4an＋2相减,得：a(n＋1)=4an－4a(n－1),即：a(n＋1)－2an=2[an－a(n－1)],则：bn=2b(n－1),其中n≥2.

a(n)>0.s(n)=[a(n)+1]/4,a(1)=s(1)=[a(1)+1]/4,a(1)=1/3.s(n+1)=[a(n+1)+1]/4,a(n+1)=s(n+1)-s(n)=[a(n+1)+

(1)Sn=(an+1)^2/4a1=S1=(a1+1)^2/4a1=14Sn=an^2+2an+14S(n-1)=a(n-1)^2+2a(n-1)+14an=(an+a(n-1))(an-a(n-1

Sn=2an-2n则Sn+1=2an+1-2（n+1）an+1=Sn+1-Sn=2an+1-2an-2则an+1-2an=2所以{an+1-2an}是等差数列（2）an+1-2an=2则an+1+2=

1、由题意,得(a1+2)/2=√(2a1)整理,得(a1-2)²=0a1-2=0a1=2(an+2)/2=√(2Sn)整理,得8Sn=(an+2)²8Sn-1=[a(n-1)+2

数列各项均为正,Sn>0.2√Sn是a(n+2)与an的等比中项,则(2√Sn)²=(an+2)an4Sn=an²+2ann=1时,4a1=4S1=a1²+2a1a1&#

1.na(n+1)=n[S(n+1)-Sn]=(n+2)SnnS(n+1)=2(n+1)SnS(n+1)/(n+1)=2*Sn/n所以{Sn/n}是公比为2的等比数列2.S1/1=a1=1所以Sn/n

题目为：a1+a2+a3=48a2*a2=a1*a4an=a1+(n-1)d求解a1、d即可带入公式可把条件转化为3a1+3d=48(a1+d)(a1+d)=a1(a1+3d)解出a1=16d=0或：

Sn+1=2Sn+n+5取n-1,得Sn=2Sn-1+n+4(在n大于等于2时）相减得a(n+1)=2an+1a(n+1)+1=2(an+1)a(n+1)+1/an+1=2an+1为等比数列.a3+1

设等差数列公差为d；a(2)^2=a(1)a(4)S(3)=48=a(1)+a(2)+a(3)=3a(2)a(2)=48/3=16；a(1)a(4)=a(2)^2=256=[a(2)-d][a(2)+

当n=1时有（a-1)a1=a(a1-1)得a1=a由（a-1)Sn=a（an-1）（a>0,n∈N*）可知：Sn=a（an-1）/（a-1)所以S(n-1)=a（a(n-1)-1）/（a-1)这里n

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+

简单的要死,你成绩在学校排中等吗?log2(Sn+1)=n,所以Sn+1=2^n,Sn=2^n-1,an=Sn-S(n-1)=（2^n-1）-（2^(n-1)-1)=2^(n-1)a(n+1)/an=

由题意得1S3=a1+a2+a3=7……1;6a2=a1+1+a3+6……22式+1式得a2=2……3将3式代入12得q=2或1/2a1=4或1an=4*(1/2)^(n-1)或an=2^(n-1)2

数列{Sn+1}是公比为2的等比数列S(n)+1=2^(n-1)(S1+1)=2^(n-1)(a1+1)①S(n-1)+1=2^(n-2)(a1+1)②①-②得an=2^(n-2)(a1+1),n≥2

1求AN的通项公式2此数列是否存在三项ar,as,at（r小于s小于t）成等差an+2为等比数列.an+2=(a1+2)2^(n-1)=2^(n+1)an=2^(n+1)

1.n=1时,S1=a1=(a1²+a1)/2,整理,得a1²-a1=0a1(a1-1)=0a1=0(与已知不符,舍去)或a1=1S1=a1=1n≥2时,Sn=(an²+

因为{Sn+1}是公比为2的等比数列,设首项为a所以Sn+1=a2^(n-1)Sn=a2^(n-1)-1n≥2时,有an=Sn-Sn-1=(a2^(n-1)-1)-[a2^(n-2)-1]=a2^(n

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