Sn=n^2an(n>=2),a1=1,求其通项公式线上等

a1=1/2,Sn=n^2an-n(n-1),求Sn和an 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 数列an满足a1=1/3,Sn=n(2n-1)an,求an a1=1,n,an,Sn成等差数列,证明｛Sn+n+2}是等比数列 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 已知数列{an}的前n项和为Sn,若a1=1/2,Sn=n^2an-n(n-1)求Sn,an 已知数列an中,a1=-2,an+1=Sn(n∈N+),求an和Sn的表达式 Sn=2an+n a1=1 求an其中{an}的前N项和为Sn Sn=na1+([n(n-1)]/2)*d=[n(a1+an)]/2【等差数列】 An=2An-1+2^n+2,n》2,A1=2,Sn为数列{An}的前N项和,证明Sn>n^3+n^2 已知a1=1,Sn=2an-2n(n>=1),则an= Sn=n(2n-1)an,a1=1/3,求an a1=1/2,an+1=an/an+2,求n/an的sn 等差数列求和公式Sn=n(a1+an)/2 或Sn=a1*n+n(n-1)d/2 注：an=a1+(n-1)d an=am+（n-m）*d（m小于n） 　　转 数列An的前n项和为Sn,已知A1=1,An+1=Sn*(n+2)/n,证明数列Sn/n是等比数列 已知数列 an前n项和为Sn,a1=1,Sn=2a(n+1),求Sn