已知数列an中4Sn=an平方+2an+1

“已知数列{an}中,an>0,Sn是数列{An}中的前n项和,且An+1/An=2Sn”An>0,求An“已知数列{an}中,an>0,Sn是数列{An}中的前n项和,且An+1/An=2Sn”An

已知数列an满足a1=1/2sn=n平方×an求an已知数列an满足a1=1/2sn=n平方×an求an已知数列an满足a1=1/2sn=n平方×an求an如图an=Sn-S(n-1)=n^2*an-

已知数列{an}中,an＞0,Sn为{an}的前n项和,且an+1/an=2Sn,求an.已知数列{an}中,an＞0,Sn为{an}的前n项和,且an+1/an=2Sn,求an.已知数列{an}中,

已知数列{an}中,an＞0,Sn为{an}的前n项和,且an+1/an=2Sn,求an.已知数列{an}中,an＞0,Sn为{an}的前n项和,且an+1/an=2Sn,求an.已知数列{an}中,

已知数列{an}中,an>o,且2Sn=an2+an,求an已知数列{an}中,an>o,且2Sn=an2+an,求an已知数列{an}中,an>o,且2Sn=an2+an,求ann=1时,2S1=2

已知数列{an}中a1=1,且满足an+an-1不等于0,Sn=1/6*(an+1)(an+2).(1)求通项an,并说明{an}是什么数列(2)求数列{an}的前n项和Sn已知数列{an}中a1=1

在数列an中,已知a1=1,Sn=n的平方*an,求通项公式an在数列an中,已知a1=1,Sn=n的平方*an,求通项公式an在数列an中,已知a1=1,Sn=n的平方*an,求通项公式anan=S

已知数列an,an属于n*,sn=1/8*(an+2)^2,{an}是等差数列已知数列an,an属于n*,sn=1/8*(an+2)^2,{an}是等差数列已知数列an,an属于n*,sn=1/8*(

已知数列an＞0an+1/an=2Sn求an已知数列an＞0an+1/an=2Sn求an已知数列an＞0an+1/an=2Sn求an因S1=a1则a1+1/a1=2a1注意到a1>0解得a1=1因Sn

数列{an}中,已知a1=1,an=2Sn^2/(2Sn-1).求an通项公式数列{an}中,已知a1=1,an=2Sn^2/(2Sn-1).求an通项公式数列{an}中,已知a1=1,an=2Sn^

已知数列an中a1=-2且an+1=sn（n+1为下标）,求an,sn已知数列an中a1=-2且an+1=sn（n+1为下标）,求an,sn已知数列an中a1=-2且an+1=sn（n+1为下标）,求

已知数列an中,a1=-2,an+1=Sn(n∈N+),求an和Sn的表达式已知数列an中,a1=-2,an+1=Sn(n∈N+),求an和Sn的表达式已知数列an中,a1=-2,an+1=Sn(n∈

已知AN中,a1=2,6Sn=（an+1)(an+2),求数列An的通项公式An和Sn已知AN中,a1=2,6Sn=（an+1)(an+2),求数列An的通项公式An和Sn已知AN中,a1=2,6Sn

已知数列an,Sn>1,6Sn=(an+1)(an+2)求通项已知数列an,Sn>1,6Sn=(an+1)(an+2)求通项已知数列an,Sn>1,6Sn=(an+1)(an+2)求通项简单啊6Sn=

已知数列an中,a1=1,当n≥2时,其前n项和Sn平方=an(Sn-1/2)求Sn表达式.已知数列an中,a1=1,当n≥2时,其前n项和Sn平方=an(Sn-1/2)求Sn表达式.已知数列an中,

已知数列{an}中,a2=2,前n项和为Sn,且Sn=n(an+1)/2证明数列{an+1-an}是等差数列已知数列{an}中,a2=2,前n项和为Sn,且Sn=n(an+1)/2证明数列{an+1-

已知数列{an}中,an>0其前n项和为Sn,且Sn=1/8(an+2)²,求证：数列{an}为等差数列已知数列{an}中,an>0其前n项和为Sn,且Sn=1/8(an+2)²,

已知数列{an}中,an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列{an}是等差数列.已知数列{an}中,an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列

数列an中,a1=1,sn+1=sn-1/2an,则an=?数列an中,a1=1,sn+1=sn-1/2an,则an=?数列an中,a1=1,sn+1=sn-1/2an,则an=?s(n+1)=sn-

已知在正整数数列｛an｝中,前n项和Sn满足：Sn=1/8(an+2)的平方(1)求证：｛an｝是等差数列(2)若bn=1/...已知在正整数数列｛an｝中,前n项和Sn满足：Sn=1/8(an+2)