S(n+1)+a1=2(Sn +a1) 数列{Sn +a1}是首项为2a1,公比为2的等比数列 从哪看出的首项和公比?

S(n+1)+a1=2(Sn +a1) 数列{Sn +a1}是首项为2a1,公比为2的等比数列 从哪看出的首项和公比? 数列{an},前n项和sn,a1=2,a1、S(n+1)、4Sn成等差数列,求{an}通项公式、Sn 等差数列公式Sn = a1+a2+...+anSn = a1+(a1+d)+(a1+d+d)+...+[a1+(n-1)d]Sn = a1*n+[1+2+...+(n-1)]*dSn = a1*n+n*(n-1)/2*dSn = a1*n+n*(n-1)*d/2Sn = (a1+an)*n/2an = a1+(n-1)*dSn = [2a1+(n-1)*d]*n/2Sn = a1*n+n*(n-1)*d/2 数列a1=3,数列满足2an=Sn乘以S(n-1),求通项公式. an=a1+(n-1)d 和Sn=n(a1+an)/2中an是什么意思? an=a1+(n-1)d 和Sn=n(a1+an)/2中an是什么 a1=1,Sn=n^2an,求an 已知数列an满足a1=1,前n项和为Sn,且Sn,S(n+1),2a1成等差数列,用数学归纳法证明：Sn=(2^n)-1/2^(n-1) a1=1/2,Sn=n^2an-n(n-1),求Sn和an 数列{an}前n项和为sn,若sn/n=3s(n-1)/n-1(n≥2),a1=3,求an 已知数列{an} 的前n项和为sn,且an=sn *s(n-1)a1=2/9 求证:{1/sn}为等差 已知Sn是数列{An}的前n项和,A1=2,根号Sn—根号S(n-1)=根号2,求Sn的表达式 Sn-S(n-1)等于an还是a1 已知等差数列an中,a1=1,前n项和Sn,若S(n+1)/Sn=(4n+2)/(n+1),求an 已知等差数列an中,a1=1,前n项和Sn,若S(n+1)/Sn=(4n+2)/(n+1),求an 数列an前n项和为sn,a1=1,2s(n+1）-sn=2.n∈n*.求an的通项公式 公式：Sn=S(n-1)+n^3*d.已知Sn和a1.怎样求得d?公式：Sn=S(n-1)*(1+n)+n^3*d 等比数列{an}的a1=1,q>0,则lim(n-->+无穷)S(n+2)/Sn=