在数列{an}中,a1=2,an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n,求证bn是等差数列,并写出其通项公式;

将an=2an-1+2^(n+1)(n>=2,)令bn=an/2^n两边除以2^n,得an/2^n=2a(n-1)/2^n+2,即
bn=a(n-1)/2^(n-1)+2,所以bn=b(n-1)+2,所以bn是等差数列.b1=a1/2=1,所以bn=2n-1