1 * 3 * 5 + 3 * 5 * 7 +. + 9 * 11 * 13 = how much,

1 * 3 * 5 + 3 * 5 * 7 +. + 9 * 11 * 13 = how much,

(2n-1)(2n+1)(2n+3)=(2n-1)(4n^2+8n+3)=8n^3+12n^2-2n-3
1+...+n=n(n+1)/2
1+...+n^2=(n+3n^2+2n^3)/6=n(n+1)(2n+1)/6
1+...+n^3=n^2(n+1)^2/4
When n = 1,2,..., K,
(2n-1) (2n + 1) (2n + 3) equals 1 * 3 * 5,3 * 5 * 7,..., (2k-1) (2k + 1) (2k + 3)
If K is 5, the sum is 1 * 3 * 5 + 3 * 5 * 7 +... + 9 * 11 * 13
8(1+...+k^3)+12(1+...+k^2)-2(1+...+k)-3k
2k^2(k+1)^2+2k(k+1)(2k+1)-k(k+1)-3k
=(k+k^2)(2k+2k^2+4k+2-1)-3k
=(k+k^2)(2k^2+6k+1)-3k
=k(2k^3+8k^2+7k+1-3)
=k(2k^3+8k^2+7k-2)
=k(k+2)(2k^2+4k-1)
K = 5, the result is
35*(70-1)=2415