an=1/(n-1)(n+2),求sn我已经知道an=1/(n-1)(n+2)=(1/(n-1)-1/(n+2)）*1/3,但是若n=1时分母为0,则无意义,

an=n(n+1) 求sn a1=1/2,Sn=n^2an-n(n-1),求Sn和an 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=(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)*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次方)*n,求sn a1=1,Sn=n^2an,求an Sn=2an+(-1)n次方 怎么求an An=1/n,求Sn. 数列an,an=(2n-1)+1/【3n(n+1)】,求Sn