a1=1,a(n+1)=(1+1/n)an+n+1/2^n,设bn＝an/n求数列bn的通项公式

A(n+1)=(1+1/n)An+(n+1)/2^n
A(n+1)=(n+1)/n×An+(n+1)/2^n
两边除n+1
A(n+1)/(n+1)=An/n+1/2^n
B(n+1)=Bn+1/2^n
Bn=B(n-1)+1/2^(n-1)
B(n-1)=B(n-2)+1/2^(n-2)
……
B2=B1+1/2
上式相加,相同项消去
Bn=B1+1/2×(1-(1/2)^(n-1))/(1-1/2)
=B1+1-1/2^(n-1)
B1=A1/1=1
Bn=2-1/2^(n-1)