：（Ⅰ）由an=f(bn)=g(bn+1)得an=4bn+1,an=2bn+1,a(n+1)=4b(n+1)+1把an=2bn+1代入得∴a(n+1)=2an+1,∴a(n+1)+1=2（an+1）,∵a1=1,∴数列{an+1}是首项为2,公比为2的等比数列．（4分）∴．an+1=2×2^(n-1)∴an=2^n-1．...