a(n+1)=[(n+1)/n]an+(n+1)/2^n两边同除(n+1)得：a(n+1)/(n+1)=an/n+1/2^n这步不懂?b1=a1/1=1b(n+1)-bn=1/2^n还有这步不懂?

Question

a(n+1)=[(n+1)/n]an+(n+1)/2^n
两边同除(n+1)得：a(n+1)/(n+1)=an/n+1/2^n这步不懂?
b1=a1/1=1
b(n+1)-bn=1/2^n还有这步不懂?

Answer 1

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