将第三个等式两边同除（n+1）得，an+1/n+1=an/n+1/2^n,由bn=an/n得，bn+1=bn+1/2^n=bn-1+1/2^n-1+1/2^n=...=b1+1/2+...+1/2^n=2-1/2^n,那么bn=2-1/2^n-1
令an=nbn带入等式得到bn+1-bn=(n+1)/2^nbn=b1+1/2^1+...+1/2^(n-1)=2*(1-(1/2)^n)
