Total no of squares on a Chess Board

Question

Is there any formula than calculates the total number of squares on chessboard?

For example in a $8\times8$ chessboard, there are squares of sizes $1\times1$, $2\times2$, $\ldots$, $8\times8$.

So I want to know their total number. Is there a general formula for a $n\times n$ chessboard?

Answer 1

There are

$1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 = 204$ squares on an $8\times8$ board.
In the general case, the number of squares is $1^2 + \dots + n^2$ for an $n\times n$ board,
or equivalently, $\rlap{\raise{2.5ex}{~\scriptsize ~n}}{\rlap{\lower{2ex}{\scriptsize{k\text{=1}}}}{\large{\sum}~k^2}}$.

As @M Oehm noted, a Faulhaber formula lets us simplify this to $\large ~~\frac{n(n+1)(2n+1)}6$.

This is because ...

For $1\times1$ squares, there are obviously $n^2$ of them.
You can fit $(n-1)$ rows of $2\times2$ squares in an $n^2$ area, and each row has $(n-1)$ in it.
You can fit $(n-2)$ rows of $3\times3$ squares in an $n^2$ area, and each row has $(n-2)$ in it.
...
You can fit $2$ rows of $(n-1)\times(n-1)$ squares in an $n^2$ area, and each row has $2$ in it.
And finally, of course, you can fit 1 $(n\times n)$ square in an $n^2$ area.

This expands to
$n^2 + (n-1)^2 + \dots + 2^2 + 1^2 = \rlap{\raise{2.5ex}{~\scriptsize ~n}}{\rlap{\lower{2ex}{\scriptsize{k\text{=1}}}}{\large{\sum}~k^2}} = \large ~~\frac{n(n+1)(2n+1)}6$.

Answer 2

In an n × n checkerboard, there are:

• n² squares of edge length 1;
• (n − 1)² squares of edge length 2;
• (n − 2)² squares of edge length 3;
• ...
• 9 squares of edge length n − 2
• 4 squares of edge length n − 1
• 1 square of edge length n

In other words, there are

   ∑ k²;   k = 1, ..., n

squares. With the Faulhaber formula, you can simplify this to:

   ¹/₆ n · (n + 1) · (2 n + 1)

On the regular checker board, where n = 8, there are 204 squares.

Caveat: This formula counts only the axis-aligned squares. If squares can be drawn with any grid point between the cells as corner, the solution is different.

Answer 3

Lets look at the patterns:

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There is $1$ square in a $1$x$1$ square.

There is $5$ squares in a $2$x$2$ square, 4 small, 1 large.

There is $14$ squares in a $3$x$3$ square, 9 small, 4 medium, 1 large

There is $30$ squares in a $4$x$4$ square, 16 small, 9 medium, 4 large, 1 extra large.

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Notice anything? These are the square pyramidal numbers according to oeis and wikipedia says

'Square pyramidal numbers also solve the problem of counting the number of squares in an n × n grid.'

For an $n$x$n$ box, it is the sum of $n^2 + (n-1)^2 + (n-2)^2 + ... + 0^2$, or more simply the sum of $n^2$ and all previous squares

From this we can derive the final formula for the total amounts of squares, $t$ in an $n$x$n$ grid:

$$t=\frac{n*(n+1)*(2*n+1)}{6}$$