n=1：1/3=1*2/2*3=1/3 设n=k,1/3 + ...+ (k^2)/(2k-1)(2k+1) = k(k+1)/2(2k+1) 成立当n=k+1：1/3+ ...+(k^2)/(2k-1)(2k+1)+(k+1)^2/(2k+1)(2k+3)=(2k+1)(k+2)(k+1)/2(2k+1)(2k+3)= (k+1)(k+2)/2[2(k+1)+1]