先证N=21/a1a2+1/a2a3=n/a1a3a3+a1=2a2 证得在N=2时1~3成立假设N=k-1成立N=k时：两式子左右相减：1/akak+1=k/a1a(k+1)-(k-1)/a1akkak-a1=(k-1)ak+1 设ak=a1+(k-1)d(ka1+k(k-1)d-a1)/(k-1)=a(k+1)=a1+kd即a(k+1)也是...