1^2+3^2+5^2+...+(2n-1)^2=1/3n(4n^2-1)
问题描述：
1^2+3^2+5^2+...+(2n-1)^2=1/3n(4n^2-1)
答
证明：因为1^2+2^2+...+n^2
=n(n+1)(2n+1)/6
所以1^2+2^2+...+(2n)^2
=2n(2n+1)(4n+1)/6
=n(2n+1)(4n+1)/3
2^2+4^2+...+(2n)^2
=4(1^2+2^2+...+n^2)
=4n(n+1)(2n+1)/6
=2n(n+1)(2n+1)/3
所以1^2+3^2+...(2n-1)^2
=[1^2+2^2+...+(2n)^2]-[2^2+4^2+...+(2n)^2]
=n(2n+1)(4n+1)/3-2n(n+1)(2n+1)/3
=n(2n+1)(2n-1)/3
=n(4n^2-1)/3