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Square Sequence (Posted on 2003-12-19) Difficulty: 5 of 5
When you add all the terms up from this sequence: x + (x-1) + 3(x-2) + (x-3) + (x-4) + (x-5) + 3(x-6) + (x-7) ... it will be equal to half of (x + x - x) for any positive even integer x. Prove why this works.

Example: 12 + 11 + 10 + 10 + 10 + 9 +8 + 7 + 6 + 6 + 6 + 5 + 4 + 3 + 2 + 2 + 2 + 1 if x = 12.

Note: The coefficients go 1, 1, then 3, then 3 1s, then 3, then 3 1s. The coefficients go in this order, even if there are coefficients left when the sequence stops. For example, with 6, the coefficients would go 1,1,3,1,1,1.

See The Solution Submitted by Gamer    
Rating: 2.5000 (4 votes)

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Solution proof | Comment 3 of 10 |
This proof is similar to Penny's, but is based on the coefficient's starting at 1,1,3,1, which they always do (once x > 3), it need be done only one way:

First, it is true for x=2, as 2^2+1^2=5 and (2^3+2^2-1)/2 = 5.

It is also true for x=4, as 4^2+3^2+3*2^2+1^2 = 38 and (4^3+4^2-4)/2 = 38.

Then, assume it is true for x-4, so that the sum is

x^2 + (x-1)^2 + 3(x-2)^2 + (x-3)^2 + ((x-4)^3 + (x-4)^2 - (x-4))/2

This expands to x^2 + x^2 - 2x + 1 + 3x^2 - 12x + 12 + x^2 - 6x + 9 + (x^3-12x^2+48x-64+x^2-8x+16-x+4)/2

This simplifies to

6x^2 - 20x + 22 + (x^3 - 11x^2 + 39x - 44)/2

which further simplifies to (x^3 + x^2 - x)/2

  Posted by Charlie on 2003-12-19 16:33:44
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