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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: x² + (x-1)² + 3(x-2)² + (x-3)² + (x-4)² + (x-5)² + 3(x-6)² + (x-7)² ... it will be equal to half of (x³ + x² - x) for any positive even integer x. Prove why this works.

Example: 12² + 11² + 10² + 10² + 10² + 9² +8² + 7² + 6² + 6² + 6² + 5² + 4² + 3² + 2² + 2² + 2² + 1² 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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): proof | Comment 6 of 10 |
(In reply to re(2): proof by Penny)

First of all, I can find nothing I said to support your claim that I was using the wrong form of inductive.

The point of this is not to say the statement's true. I have already said it is. The point is to prove it. Can you find a proof that doesn't require any induction to use?
  Posted by Gamer on 2003-12-20 07:36:11

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