1×1+3×3+5×5+.....+(2n-1)×(2n-1）=

n为正整数的情况下，
1^2+3^2+5^2+……+（2n-1）^2
=sigma（4n^2-4n+1)
=4(1^2+2^2+3^3+…+n^2)-4(1+2+3+…+n)+n
(1^2+2^2+3^2+…+n^2=[n*(n+1)*(2n+1)]/6=1/3n^3+1/2n^2+1/6n；
1+2+3+…+n＝1/2n^2+1/2n)
故1^2+3^2+5^2+……+（2n-1）^2
=4（1/3n^3+1/2n^2+1/6n）-4(1/2n^2+1/2n)+n
=4/3n^3-1/3n=1/3n(4n^2-1)