第1个回答 2019-12-17
A(n+1)=(n+1)An/n+(n+1)/2^n
A(n+1)/(n+1)=An/n+1/2^n
依此类推
An/n=A(n-1)/(n-1)+1/2^(n-1)
A(n-1)/(n-1)=A(n-2)/(n-2)+1/2^(n-2)
……
A2/2=A1/1+1/2^1
上式相加,相同项消去
An/n=A1/1+(1/2^1+1/2^2+……+1/2^(n-1))
=1+1/2×(1-(1/2)^(n-1))/(1-1/2)
=2-1/2^n
Bn=2-1/2^n