a(n+1)=(3an-2)/(2an-1) a1=2 求通项

来源：学生作业帮编辑：大师作文网作业帮分类：数学作业时间：2024/11/11 04:45:37

两边同时减去1,则:
a_{n+1}-1=(a_{n}-1)/(2a_{n}-1)
将上式取倒数,则:
1/(a_{n+1}-1)=2+1/(a_{n}-1)
1/(a_{1}-1)=1
所以,数列{1/(a_{n}-1)}是一个首项为1公差为2的等差数列.
1/(a_{n}-1)=1/(a_{1}-1)+2*(n-1)
=2n-1
a_{n}=1/(2n-1)+1=2n/(2n-1)
所以,通项为:a_{n}=2n/(2n-1)

已知数列{an}满足,a1=2,a(n+1)=3根号an,求通项an 如果数列{an}中,a1=3,a(n+1)-2an=2an*a(n+1)(an不等于0),求通项公式an 数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝. 数列{an},a1=1/2,a(n+1)=3an/(an)+3, a(n+1)=2an/3an+4,a1=1/4,求an 已知数列an中,a1=1,a(n+1)=3an+2^n,求通项公式an 数列{an},a1=1,a(n+1)=2an-n^2+3n 设数列{an}中,a1=2,a(n+1)=an+n+1,求an 数列{an},a1=2,an+1(下标)=an下标+n+1 求通项an下标数列｛an｝满足a1=1,且an=an-1+3n-2,求an 在数列an中,a1=2 an+1=an+3n则an= a1=2,a(n+1)=an^2求通项