数列满足a1=1,an+1=2n+1 （n属于正整数集）1）、求证数列{an+1}是等比数列； 2）、求｛an｝的通项公式.正确的是：数列满足a1=1，an+1=2an+1 （n属于正整数集）

数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 数列{an}满足a1=1 an+1=2n+1an/an+2n 数列an满足a1=1/3,Sn=n(2n-1)an,求an 数列{an}满足a1=2,a(n+1)=2an+n+2,求an 数列｛an｝满足a1=1,且an=an-1+3n-2,求an 已知数列{an}满足an+1=2an+3.5^n,a1=6.求an 数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010 已知数列an满足an=1+2+...+n,且1/a1+1/a2+...+1/an 已知数列{an}满足a1=1,an+1·an=2^n 则s2012 已知数列{an}满足a1=1/2,sn=n^2an,求通项an 数列[An]满足a1=2,a（n+1）=3an-2 求an 已知数列{an},满足a1=1/2,Sn=n²×an,求an 已知数列an满足a1=1/2 sn=n平方×an 求an 数列an满足a1=0,An+1=an+2n求a2009 已知数列An满足 A1=1/2 Sn=N²An 求An 已知数列an满足a1=1 Sn=2an+n 求an 数列an满足a1=1/2,a1+a2+a3……an=n^2an,则an 已知数列{An}满足A1=1,An+1=2An+2^n.求证数列An/2是等差数列