Sn=2n/n+1 ,Sn=1+1/1+2+1/1+2+3+...1/1+2+3..n这个数学归纳法当n=k+1时,Sk+1=(2k/k+1) + (1/1+2+3...

Sn=3+2^n Sn-1=3+2^(n-1).则Sn-Sn-1=? an=n(n+1) 求sn 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 2Sn+Sn-1=3-8/2^n,求Sn Sn+1=2Sn+3^n怎样通过待定系数法转化成等比数列(n、n+1下标)可以这样做吗Sn+1+K=2(Sn+k)、得到Sn+1=2Sn+K、K=3^n 已知数列 an前n项和为Sn,a1=1,Sn=2a(n+1),求Sn 正数列{bn}前n项和Sn·且Sn=1/2(bn+n/bn)求Sn an=n+2 Sn=1/a1a2+1/a2a3+.+1/ana(n+1),求Sn 已知sn=(17n-n^2)/2求：1/s1+1/s2+…+1/sn 设Sn=1*4+2*7+.n(3n+1)则Sn= Sn=n(n+2) 求1/S1+1/S2+...Sn= 已知Sn求an怎么算,例：Sn=2n^2+n+1,求通项公式 sn=1/a+2/a^2+…+n/a^n(a≠0),求sn a1=1,n,an,Sn成等差数列,证明｛Sn+n+2}是等比数列 Sn=1²+2²+3²+…+n²,用n表示Sn 证明数列是等比数列数列前n项和为Sn,a1=1,a(n+1)=（n+2)Sn/n,求证Sn/n是等比数列,