Count the number of squares present in a grid of 1 0 2 4 × 1 0 2 4 squares.
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We must count the number of 1 × 1 , 2 × 2 , ⋯ , and 1 0 2 4 × 1 0 2 4 squares. That's a lot of squares, so we should try to come up with a generalization to count the number of squares in an n × n grid.
Start by counting the number of 1 × 1 squares. There are n columns of width 1 and within each of these columns, there are n different rows of width 1 , so the number of 1 × 1 squares is n × n = n 2 .
Then count the number of 2 × 2 squares. There are n − 1 columns of width 2 and within each of these, there are also n − 1 rows of width 2 . Each 2 × 2 square is made from one of the n − 1 columns of width 2 and one of the n − 1 rows of width 2 . So, there are ( n − 1 ) × ( n − 1 ) = ( n − 1 ) 2 squares of width 2 .
Continuing the pattern, you have that the number of ( n − 1 ) × ( n − 1 ) squares is ( n − ( n − 2 ) ) × ( n − ( n − 2 ) ) = 2 2 , and the number of n × n squares is 1 2 .
Summing up the number of squares, you get: n 2 + ( n − 1 ) 2 + ⋯ + 2 2 + 1 2 = ∑ k = 1 n k 2 . In this case, n = 1 0 2 4 , so I computed ∑ k = 1 1 0 2 4 k 2 with Wolframalpha and got 3 5 8 . 4 3 8 . 4 0 0 .
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If you count the number of squares in a 1 × 1 square grid it is equal to 1 , which is equal to 1 2 .
If you count the number of squares in a 2 × 2 square grid it is equal to 5 , which is equal to 1 2 + 2 2 .
If you count the number of squares in a 3 × 3 square grid it is equal to 1 5 , which is equal to 1 2 + 2 2 + 3 2 .
If we follow this sequence we see that the number of squares in a n × n grid of squares is equal to:
1 2 + 2 2 + 3 2 + 4 2 + … + n 2 = i = 1 ∑ n i 2 = 6 n ( n + 1 ) ( 2 n + 1 ) .
1 2 + 2 2 + 3 2 + 4 2 + … + 1 0 2 4 2 = i = 1 ∑ 1 0 2 4 i 2 = 6 1 0 2 4 ( 1 0 2 4 + 1 ) ( 2 ( 1 0 2 4 ) + 1 ) ) .
= 6 1 0 2 4 × 1 0 2 5 × 2 0 4 9
= 3 5 8 4 3 8 4 0 0 .