I know that you believe you understand what you think I said,
but I'm not sure you realize that what you heard is not what I
About -fake made up name here-.
And ha ha ha. 1,2,4,?,16,32,64?
Hm no. It should be more like, let's see. Oh jesss!
Show that the number of squares whose vertices lie at the centers of
the squares of an n by n checkerboard is given by n2(n2 - 1)/12.
As an example, for the case 4 by 4, there are:
* 9 squares of size 2 x 2 (like the green one)
* 4 squares of size 3 x 3 (like the blue one)
* 1 square of size 4 x 4 (like the red one)
* 4 diamonds (like the yellow one)
* 2 skewed squares (like the purple one)
for a total of 20.
If we consider the number of squares that have their vertices on the
border cells of an k by k grid, then the number is clearly (k-1). For
example, there are three such squares on a 4 by 4 grid:
Within an n x n grid, we have:
* (n-1) squares based on the border of the n x n grid itself
* (n-2) squares based on the border of each (n-1) by (n-1)
subgrids, of which there are 4
* (n-3) squares based on the border of each (n-2) by (n-2)
subgrids, of which there are 9
* 1 square based on the border of each 2 by 2 subgrid, of which
there are (n-1)2
The total number of squares is thus:
(n-1)12 + (n-2)22 + ... + 2(n-2)2 + 1(n-1)2
= Σ k=1,2..(n-1) (k(n-k)2)
= Σ k=1,2..(n-1) ((n-k)k2)
= Σ k=1,2..(n-1) (nk2 - k3)
= n[n(n-1)(2n-1)/6] - n2(n-1)2/4
= n2(n-1)[(2n-1)/6 - (n-1)/4]
And that is just some 9th grade Geometry, good lord! Simple. XDD
I make myself laugh.
And the numbers would change some every day, giving a different
answer. Maybe it being 9th grade is too kind?
If this was the case, nobody would feel like making new accounts. Ha
ha ha. I mean, would you go that far? Maybe for your friends here,
yes, but to be fake? Nah.