Sn=1/2n∧2+1/2n 求sn/s(n+1)

Sn=1/2n∧2+1/2n 求sn/s(n+1) 已知数列{an}中,a1=1,前n项和Sn满足Sn根号(Sn-1) -Sn-1根号(Sn)=2根号(SnSn-1)(n>=2)求an√SnS(n-1)*［√Sn-√S(n-1)］=2√SnS(n-1)√Sn-√S(n-1)=2若√Sn=0呢? 两个等差数列{an},(bn}前n项和分别为Sn,S'n,若Sn/S'n=(2n+3)/(3n-1)求a9/b9 a1=1/2,Sn=n^2an-n(n-1),求Sn和an 2Sn+Sn-1=3-8/2^n,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 等差数列题目S(n+1)/Sn=(n+2)/n,求An 数列{an}前n项和为sn,若sn/n=3s(n-1)/n-1(n≥2),a1=3,求an 设Sn=1+2+3...+n,f(n)=Sn/[(n+32)S(n+1)]的最大值为? 已知Sn是数列{An}的前n项和,A1=2,根号Sn—根号S(n-1)=根号2,求Sn的表达式 数列{an},前n项和sn,a1=2,a1、S(n+1)、4Sn成等差数列,求{an}通项公式、Sn 高二数学数列求和问题!在数列an中,a1=1,Sn为an前n项和,an=S(n-1) 求Sn,an请帮我看看我分别先算Sn和an哪里出错了?1.先算Sn因为an=2Sn-1 且an=Sn-Sn-1所以 Sn-Sn-1=2Sn-1Sn=3Sn-1所以Sn是一个等比数列 公比为3 运 an=(2^n+1/(2^n))^2求Sn an=(2n-1)*2^(n-1),求Sn an=(2n-1)2^n-1,求sn? 已知数列an中,a1等于1,当n大于等于2,其前n项和Sn满足Sn的平方等于an乘以(Sn-1/2),求(1)Sn的表达式(2)设bn等于Sn/2n+1,求{bn}的前n项和Tn第1问搞好了 n>=2时，Sn^2=(Sn-S(n-1))(Sn-1/2) 化简得0=-SnS(n-1)-(1/2)Sn+(1/2 an=(2n+1)*2^n,求Sn 已知an=1/2n(n+1),求Sn