数列{an}满足an+1=1/(2-an),
一直数列{an}满足a1=0,an=(an-1 +4)/(2an-1) ,求 an
若数列{An}满足An+1=An^2,则称数列{An}为“平方递推数列”,已知数列{an}中,a1=9,点(an,an+
已知数列{an}满足a1=1,an+1=2an+1 1)求证:数列{an+1}为等比数列; 2) 求{an}的通项an
已知数列{An}满足:A1=3 ,An+1=(3An-2)/An,n属于N*.1)证明:数列{(An--1)/(An--
已知数列{an}满足:a1=3,an+1=(3an-2)/an ,n∈N*.(Ⅰ)证明数列{(an-1)/an-2
已知数列{an}满足an+1=2an+n+1(n∈N*).
数列{an}满足a1=1 an+1=2n+1an/an+2n
数列{an}满足a1=1,且an=an-1+3n-2,求an
数列an满足a1=2,an+1=4an+9,则an=?
已知数列{an}满足an+1=2an+3.5^n,a1=6.求an
数列an满足a1=2,an+1=an²求an
已知数列an满足条件a1=-2 a(n+1)=2an/(1-an) 则an=