2^2+4^2+6^2+…+(2n)^2=2/3n(n+1)(2n+1).

2^2+4^2+6^2+…+(2n)^2=2/3n(n+1)(2n+1).
2^2+4^2+6^2+…+(2n)^2=2/3n(n+1)(2n+1).

2^2+4^2+6^2+…+(2n)^2=2/3n(n+1)(2n+1).
证明： (1)当n=1时,左边=4,右边=4,等式成立. (2)假设n=k时等式成立,即 2+4+6+…+(2k)=（2/3)k(k+1)(2k+1). 那么, 2+4+6+…+(2k)+[2(k+1)]=(2/3)k(k+1)(2k+1)+[2(k+1)] =(2/3)k(k+1)(2k+1)+4(k+1)(k+1) =2(k+1)[(1/3)k(2k+1)+2(k+1)] =2(k+1)(1/3)(2k+k+6k+6) =(2/3)(k+1)(2k+7k+6) =(2/3)(k+1)(k+2)(2k+3) 补充： 也就是说,当n=k+1时等式也成立. 根据(!)和(2),可知等式对任何n∈ 正整数 都成立