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1^2 +2^2 +3^2+.... + 8^2= 8*9*17/6=204
@directrix, u know the answer?
(n(n+1)(2n+1))/6 , let n = 8
@Arnabo09 Why do you ask?
then tell me if its correct or not.. i guess u know the answer
I would like to get input from everyone who wants to reply. By then, I will have finished counting on my checkerboard. :)
what i also have found is that a similar approach works for an (albeit imaginary) 3D "chess cube" the sum of the cubes: (n(n+1)/2)^2
oh.. ok.. there will be 8 squares in a row if its length is 1 unit 7 squares in a row if its length is 2 unit.. 1 square in a row if its length is 8 unit.. so, total: 1^2+2^2+3^2+..+8^2
what happens if we had a chessboard which isnt square? instead of NxN , we had NxM ?
well, i guess if chessboard is mentioned, it is assumed its 8x8
overlaping or not?
well the problem is solved, i was just suggesting an extension to it
if overlaping: 1+2*2+3*3+4*4+5*5+6*6+7*7+8*8
then the numbers shouldnt be squared.. say, if its 8x6 1 unit length: 8*6 2 unit length:7*5 6 unit length: 3*1.. add these up, @eigen..
if they are alowed to overlap each other, my answer is correct
if you whant i explain....
if not overlaping, you are right
204. There are 8 different sizes of squares, ranging from 1 × 1 to 8 × 8. Only 1 square is 8 × 8, but 4 squares are 7 × 7 and 9 squares are 6 × 6; this pattern continues through to 64 1 × 1 squares.
I think it is 1 + 4+9+16+25+36+49 +64 ..
Oh fail, I'm too late :S