a1=1/2,Sn=n^2an-n(n-1),求Sn和an

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 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an满足a1=1/3,Sn=n(2n-1)an,求an Sn=na1+([n(n-1)]/2)*d=[n(a1+an)]/2【等差数列】 a1=1,n,an,Sn成等差数列,证明｛Sn+n+2}是等比数列 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 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 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 ,a1=1,前n项和为Sn ,正整数n对应的n an Sn 成等差数列.1.证明{Sn+n+2}成等比数列,2.求{n+2/n(n+1)(1+an)}前n项和 已知a1=1,Sn=2an-2n(n>=1),则an= Sn=n(2n-1)an,a1=1/3,求an 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 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