已知数列an满足a1=1.a2=3,an+2=3an+1-2an

a(n+2)=3*a(n+1)-2*an
a(n+2)-a(n+1)=2*(a(n+1)-an)
a2-a1=3-1=2
a(n+1)-an=2^n
a(n+2)-2a(n+1)=a(n+1)-2*an
a2-2*a1=3-2=1
a(n+1)-2*an=1
an=2^n-1
4^(b1-1)*4^(b2-1)*…*4^(bn-1)=4^(b1+b2+…+bn-n)
an+1=2^n
4^(b1+b2+…+bn-n)=(an+1)^bn=4^(n*bn/2)
b1+b2+…+bn=(n/2)*bn+n
b1+b2+…+b(n-1)=((n-1)/2)*b(n-1)+(n-1)
bn/(n-1)=b(n-1)/(n-2)-2/((n-2)(n-1))
b2/1=3
bn/(n-1)=(n+1)/(n-1)
bn=n+1
故bn为等差数列