a(n+1)=[(n+1)/n]an+(n+1)/2^n,a(n+1)/(n+1)=[(n+1)/n]*1/(n+1)*an+(n+1)/2^n*1/(n+1),约分（n+1）得：a(n+1)/(n+1)=an/n+1/2^n. 令bn=an/n,则b(n+1)=a(n+1)/(n+1).故b1=a1/1=1,b(n+1)-bn=1/2^n