数列{an}满足a1=2/3,an+1-an=√2/3(an-1+an） 求an,有严谨的做法么老师讲解都是用猜想的做法,求出a1a2来找规律的,那种做法我不需要哦根号包括全部.而而且题目根号里面的是an+1+an.疏忽啦

数列｛an｝满足a1=1,且an=an-1+3n-2,求an 数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 已知数列{an}满足a1=1 an+1=an/(3an+1) 则球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=1/2,an+1=3an/an+3求证1/an为等差数列 数列an满足a1=1/2 an+1=an/(2an+3) 猜想数列通项公式 已知数列{an}满足a1=1/2,an+1=3an+1,求数列{an}通项公式 数列{an}满足a1=a,an+1=1+1/an.若3/2 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）=3an-2 求an 已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an 数列an满足a1=1,a2=1/3,并且an(an-1+an+1)=2an+1an-1.则数列第2012项为? 数列{an}满足a1=1 an+1=2n+1an/an+2n 数列an满足a1=2,an+1=4an+9,则an=? 数列[An]满足An+1-An+3=0,且A1=-5.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=2,an+1=an²求an 数列an满足a1=2,an+1=an²求an 设数列AN满足A1等于1,3(A1+a2+~+AN)=(n+2)an,求通向公式