(1)B(1)=A(1)/1=1
由A(n+1)=(1+1/n)A(n)+(n+1)/2^n=(n+1)A(n)/n+(n+1)/2^n
A(n+1)/(n+1)=A(n)/n+1/2^n
即B(n+1)=B(n)+1/2^n
B(n)=B(n-1)+1/2^(n-1)
=B(n-2)+1/2^(n-2)+1/2^(n-1)
=…
=B(1)+1/2+…+1/2^(n-1)
=1+1/2+…+1/2^(n-1)
=2-1/2^(n-1);
(2)A(n)=nB(n)=2n-n/2^(n-1)
S(n)=2(1+2+…+n)-[1+2/2+3/4+…+n/2^(n-1)]=n^2+n-T(n)
其中T(n)=1+2/2+3/4+4/8+…+n/2^(n-1)
2T(n)=2+2/1+3/2+4/4+…+n/2^(n-2)
兩式相減T(n)=2+1+1/2+1/4+…+1/2^(n-2)-n/2^(n-1)=4-(n+2)/2^(n-1)
S(n)=n^2+n-4+(n+2)/2^(n-1).