Sn=(n+n^2)/2,如果用裂项法求和怎么求?RT~学了好久了,都忘记了.= =如果用裂项求1/S1+1/S2+……+1/Sn,怎么求.有什么通式么?

an=n*2^n,求Sn an=2^n+n,求Sn 已知等差数列{an}的前n项和为Sn,如果Sn=(an+1/2)^2(n∈N+0,bn=(-1)^n*Sn已知等差数列{an}的前n项和Sn,如果Sn=(an+1/2)^2(n∈N+0,bn=(-1)^n*Sn,试求{bn}的前n项和Tn 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 Sn=1/2n∧2+1/2n 求sn/s(n+1) a1=1/2,Sn=n^2an-n(n-1),求Sn和an 等差数列{an}中Sn=n^2-15n则使Sn为最小值n是 已知sn=32n-n^2求等差数列|an|的前n项和sn 已知等差数列前几项之和Sn=n^2-17n,则使Sn最小的n等于? 已知等差数列前几项之和Sn=n^2-17n,则使Sn最小的n等于? 已知等差数列前几项之和Sn=n^2-17n,则使Sn最小的n等于 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? 如果数列{a}的前项和分别为(1)Sn=3n²-2n,(2)Sn=10^n+1,求{a}的通项公式 Sn=3+2^n Sn-1=3+2^(n-1).则Sn-Sn-1=? 问一道极限化简题Sn=n^2an,n→+∞,an=2(1/n-1/(n+1)) limSn=n^2*2(1/n-1/(n+1)) =2（n^2/n-n^2/(n+1)) =2(n-n/(1+1/n)) =2(n-n) =0这样做对吗,如果不对,请说明理由,并给出正确答案 sn=2*3^n,求an an=(2^n+1/(2^n))^2求Sn 已知数列an=n^2-n+2,求Sn