an=a1+(n-1)d 和Sn=n(a1+an)/2中an是什么

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=1/2,Sn=n^2an-n(n-1),求Sn和an 已知数列 an前n项和为Sn,a1=1,Sn=2a(n+1),求Sn 数列{An},A1=1,A(n+1)=3An+4.求An和Sn. 已知数列｛an｝满足a1=1,a（n+1）=sn+（n+1）求：an和sn an是等差数列,bn满足bn=an*a(n+1)*a(n+2),bn的前n项和是Sn,若a1=d,用数学归纳法证明Sn=bn*a(n+3)/4d. 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 在等比数列An中,a1=2,前N项和为Sn,若数列An + 1 也是等比数列 ,则SN等于?A 2^（n+1） -2 B 3^n C 2^n D 3^n-1 在数列{an}中,a1=2,sn=4A(n+1) +1 ,n属于N*.求数列{an}的前n项和Sn 在等差数列an中.a1=1,Sn=a1+a2+...+an,求lim(1+Sn)/(n(1-a(n+1))),d不等于0 等差数列前n项和为Sn,若a（n+1）=3Sn,a1=1,则通项an? 设数列{an}的前n项和为Sn,a1=10,a(n+1)=9Sn+10 设数列an的前n项和为Sn,已知a1=1,(2Sn)/n=a(n+1)-1/3n^2-n-2/3 Sn=na1+([n(n-1)]/2)*d=[n(a1+an)]/2【等差数列】 等差数列求和公式　Sn=(a1+an)n/2　Sn=n(2a1+(n-1)d)/2; d=公差　Sn=An2+Bn; A=d/2,B=a1-(d/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 已知等差数列中,d=2,an=1,sn=-8,求a1和n.