(1)a(n+1)/2^(n+1) =an/(an+2^n)2^(n+1)/a(n+1)=(an+2^n)/an =1+2^n/an2^(n+1)/a(n+1)-2^n/an=1所以{2^n/an}是以公差d=1的等差数列(2)2^n/an=a1+(n-1)d =1+n-1=n所以an=2^n/n(3)bn=n(n+1) an =(n+1)2^n Sn=b1+b2+......=-[2+2^2+2^3+...+2^n]+(n+1)*2^(n+1)-2=- 2(1-2^n)/(1-2) +(n+1)*2^(n+1)-2=2-2^(n+1)+(n+1)2^(n+1)-2=n*2^(n+1)