俩边同时除以(n+1) 即a(n+1)/(n+1)=an/n+1/2^n 所以b(n+1)-bn=1/2^n 所以bn-b1=sum(k=2 to n)1/2^(k-1)=1-1/2^(n-1) 所以bn=2-1/2^(n-1) an=2n-n/2^(n-1) Sn=n(n+1)-sum(k/2^(k-1))=n(n+1)-sum(k=1 to n)((2k+2)*1/2^(k-1)-(2k+4)*1/2^k)=n(n+1)-(4-(2n+4)/2^n=n(n+1)+(n+2)/2^(n-1)-4
