Let be a positive integer. Look at the first row of larger squares in the complement of the blue grid you get from colouring squares corresponding to multiples of
blue. First, we are going to prove that the numbers within the
larger square from the left add to
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where is the
triangular number.
Let’s call the square in question First note that the numbers within the top row of
are
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Adding those up we get the sum
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which we can rearrange to give
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(1) |
As we mentioned in the main article, there’s a formula for the sum of the first integers:
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(2) |
Substituting this into the left part of expression (1) gives
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(3) |
Now the triangular number is equal to the sum of the first
integers, so it’s also equal to the right hand side of equation (2). This means that expression (3) is equal to
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Now notice that the numbers in the second row of are 2 times the numbers in the first row, so their sum is equal to
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The numbers in the second row of are 3 times the numbers in the first row, so their sum is equal to
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We can continue in this vein until we come to the sum of the last row of , which is equal to
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Adding up the sums of all the rows in therefore gives
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(4) |
Factoring out the transforms this sum to
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(5) |
Now the second bracket here is equal to , the
triangular number. Therefore our sum (5) is equal to
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as we claimed.