k>0正数 k+1,k+2,k+3,...K+99,k+100 前50个偶数的和减去前50个奇数的和的差:s=(k+2)+(k+4)+(k+100)-[(k+1)+(k+3)...+(k+99)] =[(k+2)-(k+1)]+[(k+4)-(k+3)+...+[((k+100)-(k+99)] =1+1+...+1 =50