An $n \times n$ chessboard is made up of $n$ adjacent units on either side. Here, unit means side of each small square.
Now, $n$ adjacent units on horizontal side and $n$ adjacent units on vertical side make a square of $n \times n$ dimension.
Note: The horizontal and vertical sides must have common originating point.
Therefore, for a $n x n$ chessboard, following observations can be made-
Dimension i |
No. of Horizontal Sides of i Adjacent Units |
No. of Vertical Sides of i Adjacent Units |
Total No. of Squares |
$n \times n$ |
$n - n + 1 = 1$ |
$n - n + 1 = 1$ |
$1^{2}$ |
$(n-1) \times (n-1)$ |
$n - (n-1) + 1 = 2$ |
$n - (n-1) + 1 = 2$ |
$2^{2}$ |
$(n-2) \times (n-2)$ |
$n - (n-2) + 1 = 3$ |
$n - (n-2) + 1 = 3$ |
$3^{2}$ |
... |
... |
... |
... |
$3 \times 3$ |
$n - 3+ 1 = (n-2)$ |
$n - 3+ 1 = (n-2)$ |
$(n-2)^{2}$ |
$2 \times 2$ |
$n - 2 + 1 = (n-1)$ |
$n - 2 + 1 = (n-1)$ |
$(n-1)^{2}$ |
$1 \times 1$ |
$n - 1 + 1 = n$ |
$n - 1 + 1 = n$ |
$n^{2}$ |
Therefore, total no. of squares = $1^{2} + 2^{2} + 3^{2} + ... + (n-2)^{2} + (n-1)^{2} + n^{2} = \frac{(n)(n+1)(2n+1)}{6}$
Therefore, total no. of squares in 8 x 8 chessboard = $\frac{(8)(8+1)((2 \times 8)+1)}{6} = 204$