Sn=1+(1+1/n)+(1+2/n)+···+[1+(n-1)/n] 求数列前n项和

an是等差数列,求​lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)/2n²=1+2/n+1/n²我就想知道第一步怎么来的 an=n(n+1) 求sn Sn=1/2n∧2+1/2n 求sn/s(n+1) a1=1/2,Sn=n^2an-n(n-1),求Sn和an Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗? Sn=3+2^n Sn-1=3+2^(n-1).则Sn-Sn-1=? an=(2n-1)*2^(n-1),求Sn an=(2n-1)2^n-1,求sn? 已知an=1/2n(n+1),求Sn an=(2^n+1/(2^n))^2求Sn an=(2n-1)3^n,Sn=? an=(2n+1)*2^n,求Sn 知Sn=n²+2n+1求an 已知an=(2n+1)*3^n,求Sn An=n×2^(n-1),求Sn an=（2^n-1）n,求Sn 已知:Sn=n平方+2n+1,求An an=1/【n*(n+2) 】sn极限