a(n+1)=(n+1)/n*an+(n+1)/2^n邻边除以n+1a(n+1)/(n+1)=an/n+1/2^n即b(n+1)-bn=1/2^n所以bn-b(n-1)=1/2^(n-1)……b2-b1=1/2相加bn-b1=1/2*[1-1/2^(n-1)]/(1-1/2)=1-1/2^(n-1)b1=a1/1=1所以bn=2-1/2^(n-1)...