a(n+1)=(3an-2)/(2an-1) a1=2 求通项

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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)

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