If we have, for even m,

m/2 black squares of
2*2
4*4
6*6
8*8
...
m*m

and m/2 black rectangles of
1*(m-1)
3*(m-3)
5*(m-5)
7*(m-7)
..
(m-1)*1

and we have

m/2 white squares of
1*1
3*3
5*5
7*7
...
(m-1)*(m-1)

and m/2 white rectangles of
2*m
4*(m-2)
6*(m-4)
8*(m-6)
..
m*2

then we have the same total area for all of the white
squares/rectangles as for the black squares/rectangles (which is
easily shown using algebra).

My question is, which I have not been able to do yet myself (though I
have not tried really hard),

can all of the squares/rectangles be arranged in some way so that it
becomes visually obvious that the total areas between the black and
white are the same for every even m?

( This is similar to the "Proof Without Words" in the MAA's
Mathematics Magazine.)

thanks,
Leroy Quet