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.

Mathematical induction to the rescue !!!!!

Assume, for positive even integer x (such as 6) with series coefficients ending with (3,1,1,1), that:

x^2 + (x-1)^2 +3(x-2)^2 + (x-3)^2 + (x-4)^2 + (x-5)^2 +3(x-6)^2 + (x-7)^2 +...
+ 3(4^2) + 3^2 +2^2 + 1^2 = (1/2)(x^3 + x^2 -x)

Then x+4 will also be a positive even integer with coefficients (3,1,1,1).

the series for x+4 is:

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

But by our assumption for x, this becomes:

= [(x+4)^2 + (x+3)^2 +3(x+2)^2 + (x+1)^2] + [(1/2)(x^3 + x^2 - x)]

=(1/2)(x^3 + 13x^2 + 55x + 76)

= (1/2)[x^3 + 8x^2 + 16x + 4x^2 + 32x + 64) + x^2 + 8x + 16 - x - 4]

= (1/2)[(x+4)^3 + (x+4)^2 - (x+4)]

So we have proved that the theory is true for integers 6, 10, 14, 18, 22,.......

The proofs for positive even integers with series coefficients ending with (1,3,1,1), (1,1,3,1) and (1,1,1,3) are almost identical to the foregoing. I will omit those details.

Edited on December 19, 2003, 4:02 pm

Posted by Penny on 2003-12-19 15:33:16