## Trivia

Alex Y

Okay, stop reading if you are looking for a clever puzzle for entertainment--you will probably find this boring. You have been warned...

We have expressed the sum of squares as the total of numbers in a triangle made up of one 1, followed by two 2s, then three 3s.... n ns. For illustration, I'm going to use n=5. If x is the sum of squares, then x is the sum of numbers in this triangle (imagine it as equilateral):

1

2 2

3 3 3

4 4 4 4

5 5 5 5 5

Now rotate this triangle (120 degrees if you have it as equilateral) twice to get triangles with the same elements:

5

4 5

3 4 5

2 3 4 5

1 2 3 4 5

and

5

5 4

5 4 3

5 4 3 2

5 4 3 2 1

now, 3x is gotten by adding these three triangles. In our case, the sum in each position is 11, which is 2n+1 (this holds for all n).

We have 1 + 2 + 3 + 4 + 5 = 15 = n(n+1)/2 positions

So the total of the numbers in these three triangles is

(2n+1)*n*(n+1)/2

This is the sum of all three triangles, which is three times the sum of squares we want, so x = (2n+1)*n*(n+1)/6, or with a little manipulation,

(n^3)/3 + (n^2)/2 + n/6

Testing for n=5:

125/3 + 25/2 +5/6 = (250+75+5)/6 = 330/6 = 55

1+4+9+16+25 = 55

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