a(n+1)=2an+2^n,bn=an/2^(n-1),b(n+1)=a(n+1)/2^n,b1=a1/2^0=1a(n+1)/2^n=an/2^(n-1)+1,b(n+1)=bn+1,bn为首项为1公差为1的等比数列,bn=nan=n*2^(n-1),a1=1,s1=1sn-s(n-1)=n*2^nsn/2^n-s(n-1)/2^n=nsn/2^n-1/2*s(n-1)...