在数列A（n）中,A（1）=1,A（n+1）=（1+1/n）An+（n+1）/2^n设Bn=An/n,求Bn的通项公式求数列An的前n项和Sn

A(n+1)=An+An/n+1/2^n+n/2^n,∴(n+1)Bn+(n+1)/2^n=A(n+1)∴B(n+1)=Bn+1/2^n∴∑｛B（n+1)-Bn}=∑1/2^n，故Bn=B1+∑1/2^(n-1),即Bn=2-1/2^(n-1)
An=2n-n/2^(n-1),然后错位相减
Sn=n^2+n-4+(n+2)/2^(n-1)

(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)
两式相减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).