在数列{an}中，a1=1,an+1=(1+1⼀n)an+(n+1)⼀2^n(1)设bn=an⼀n求数列{bn}的通项公式（2）求数列{an}的前n项和

(1) ∵an+1=(1+1/n)an+(n+1)/2^n=(n+1)/n*an+(n+1)/2^n
∴an+1/(n+1)=an/n+1/2^n
∵bn=an/n，∴bn+1=bn+1/2^n
bn=bn-1+1/2^(n-1)
bn-1=bn-2+1/2^(n-2)
......
b2=b1+1/2^1
b1=a1/1=1
将上述n个式子加起来，得
bn=1+1/2+1/2^2+...+1/2^(n-1)
=1+1/2(1-(1/2)^(n-1))/(1-1/2)
=1+1-1/2^(n-1)
=2-1/2^(n-1)
(2) ∵bn=an/n，∴an=n*bn
∴Sn=1*b1+2*b2+...+n*bn
=1*(2-1)+2*(2-1/2)+3*(2-1/4)+...+n*(2-1/2^(n-1))
=2(1+2+3+...+n)-(1+2/2+3/4+...+n/2^(n-1))
=2Pn-Qn
设Pn=(1+2+3+...+n)=n(n+1)/2 Qn=(1+2/2+3/2^2+4/2^3+...+n/2^(n-1))
对Qn，有 Qn/2=(1/2+2/2^2+3/2^3+...+(n-1)/2^(n-1)+n/2^n)
Qn-Qn/2=Qn/2=1+[1/2+1/2^2+1/2^3+...+1/2^(n-1)]+n/2^n
=1+1/2(1-(1/2)^(n-1))/(1-1/2)+n/2^n
=1+1-(1/2)^(n-1)+n/2^n
=2-1/2^(n-1)+n/2^n
=2+(n+2)/2^n
∴Qn=4+2(n+2)/2^n
∴Sn=2Pn-Qn
=2*n(n+1)/2-4-2(n+2)/2^n
=n(n+1)-4-2(n+2)/2^n

a(n+1)=(1+1/n)an+(n+1)/2^n
两边同除以n+1,得a(n+1)/(n+1)=an/n+1/2^n
因为bn=an/n
b(n+1)=bn+1/2^n
所以b(n+1)-b1=1/2+1/4+1/8+1/16+....+1/2^n
b1=a1=1
所以b(n+1)=1+1/2+1/4+...+1/2^n
所以bn=1+1/2+1/4+..+1/2^(n-1)=2*【1-（1/2）^n】
即bn=2-2^(1-n)

2、bn=an/n
则an=nbn
an=[2-2^(1-n)]n=2n-n*2^(1-n)

an=[2-2^(1-n)]n=2n-n*2^(1-n)