已知数列{an}中,a1=2,an+1=4an-2/3an-1 bn=3an-2/an-1 求证；数列{bn}是等比数列,并求{an}的通项公式

1.
bn=(3an-2)/(an-1)
an=(bn-2)/(bn-3)
a(n+1)=[b(n+1)-2]/[b(n+1)-3]
a(n+1)=(4an-2)/(3an-1)
3a(n+1)an-a(n+1)=4an-2
3{[b(n+1)-2]/[b(n+1)-3]}[(bn-2)/(bn-3)]-[b(n+1)-2]/[b(n+1)-3]=4(bn-2)/(bn-3)-2
3[b(n+1)-2](bn-2)-[b(n+1)-2](bn-3)=4(bn-2)[b(n+1)-3]-2[b(n+1)-3](bn-3)
2b(n+1)bn-3b(n+1)-4bn+6=2b(n+1)bn-2b(n+1)-6bn+6
b(n+1)=2bn
bn=b1*2^(n-1)
b1=(3a1-2)/(a1-1)=4
bn=b1*2^(n-1)=2^(n+1)为比数列；
2.
bn=(3an-2)/(an-1)=2^(n+1)
3an-2=[2^(n+1)]an-2^(n+1)
[3-2^(n+1)]an=2-2^(n+1)
an=[2-2^(n+1)]/[3-2^(n+1)]