还是我理解错了.已知数列｛an｝满足a1=m（m为正整数）,an+1=an/2(an为偶数）或an+1=3an+1（an

已知数列an满足 a1=1/2,an+1=3an/an+3求证1/an为等差数列 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 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 n为奇数,an+2,n为偶数},且a1+a3+a5+.+a2k--=3 已知数列｛an｝满足a1=1,an+1=2an/（an+2）（n∈N+）,则数列｛an｝的通项公式为 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a1=2,an+1=2an/an+2.求证数列{1/an}是否为等差数列 并求出an 已知数列（an）满足an+1=2an+3×2∧n,a1=2求an 数列{an}满足a1=3/2,an+1=an^2-an+1(n属于正整数),则m=1/a1+1/a2+……+1/a200 已知数列{an}满足an+1=(3an+1)/(an+3),a1=-1/3 求证1/(an)+1为等差数列,求an 已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an