2^2+4^2+6^2+…+(2n)^2=2/3n(n+1)(2n+1).
问题描述：
2^2+4^2+6^2+…+(2n)^2=2/3n(n+1)(2n+1).
答
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...