Question

数学归纳法证明:2^2+4^2+..+(2n)^2=2n(n+1)(2n+1)/3

Answer 1

证明：

(1)当n=1时,2^2=2*（1+1）*（2+1）/3=4,等式显然成立

（2）假设n=k时,等式成立,即2^2+4^2+.+(2k)^2=2k(k+1)(2k+1)/3

当n=k+1时,

2^2+4^2+.+(2k)^2+(2k+2)^2

=2k(k+1)(2k+1)/3+(2k+2)^2

=4/3*k^3+2k^2+2/3*k+4k^2+8k+4

=2/3*(2k^3+3k^2+6k^2+9k+4k+6)

=2/3*[2k(k^2+3k+2)+3(k^2+3k+2)]

=2/3*(k^2+3k+2)(2k+3)

=2/3(k+1)(k+2)(2k+3)

所以当n=k+1时等式也成立

综上,可知对任意正整数n,等式2^2+4^2+..+(2n)^2=2n(n+1)(2n+1)/3均成立