a(n+1)=[2^(n+1)a(n)]/(a(n)+2^n)a(n+1)/2^(n+1)=a(n)/(a(n)+2^n)两边倒数2^(n+1)/a(n+1)=1+2^n/a(n) 2^1/a(1)=2所以{2^n/a(n)}是首项为2，公差为1的等差数列，2^n/a(n)=2+n-1=n+1a(n)=2^n/(n+1)b(n)=2^nS(n)=2(2^n-1)
