证明：
（1）令n=1,(2×1)²=(2/3)×1×(1+1)×(2×1+1)成立；
（2）假设n=k(1≤k∈Z),等式成立,
即2²+4²+6²+…+(2k)²=(2/3)k(k+1)(2k+1)；
则当n=k+1时,
2²+4²+6²+…+(2k)²+[2(k+1)]²
=(2/3)k(k+1)(2k+1)+[2(k+1)]²
=(2/3)(k+1)[k(2k+1)+6(k+1)]
=(2/3)(k+1)[2k²+7k+6)]
=(2/3)(k+1)(k+2)(2k+3)
=(2/3)(k+1)(k+2)[2(k+1)+1]
等式也成立,得证!