Question

a(n+1)=(3an-2)/(2an-1) a1=2 求通项

Answer 1

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)