设bn=an/2^nb(n+1)=a(n+1)/2^(n+1)b(n+1)-bn=a(n+1)/2^(n+1)-an/2^n=[a(n+1)-2an]/2^(n+1)=[a(n+1)-an-an]/2^(n+1)=(Δan-an)/2^(n+1)=2^n/2^(n+1)=1/2也就是说数列b（n+1)与bn的差值是定值 即公差为1/2所以数列bn为...