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.

How to derive the formula?
Well, the series looks to be the sum of two separate series.
1) The sum of the first x squares
2) The ‘added’ terms, (due to the 3 coefficients) which are
2 x (x-2)^2
2 x (x-6)^2
………..
The sum of the first x squares is given by,
[x(x+1)(2x+1)]/6
(do we need to derive this??)

The function for the ‘added’ terms become clearer when written out,

X= sum of ‘added’ terms
2 0
4 8
6 32
8 80
10 160
12 280
. .
. .
dividing the sum of ‘added’ terms by 8 we realize they are the sum of the first triangular numbers,

X= sum of ‘added’ terms/8
2 0
4 1 (the sum of the first tri’ number)
6 4 (the sum of the first 2 tri’ numbers)
8 10 (the sum of the first 3 tri’ numbers)
10 20 (the sum of the first 4 tri’ numbers)
12 35 (the sum of the first 5 tri’ numbers)

So to complete the sum we need to add (to the simple ‘sum of the first x squares’) “8 times the sum of the first (x-2)/2 triangular numbers.”

The sum of the first n triangular numbers is given by

n (n+1)(n+2)/6
and with (x-2)/2 for n and multiplying by 8 gives,
[8 x ((x-2)/2) x (((x-2)/2) + 1) x (((x-2)/2) +2)]/6
which reduces to
[x^3 – 4x]/6

adding the two expressions together
[the sum of the first x squares] + [the ‘added’ terms]
[x(x+1)(2x+1)]/6 + [x^3 – 4x]/6
or
[3x^3 + 3x^2 – 3x]/6
cancel to give
[x^3 + x^2 – x]/2
which is the final expression for the series and was to be shown.

[Note x is required to be even since ((x-2)/2) is required to be an integer (or 0)]
I know it's not as robust as induction, it's just how I looked at it. Go easy

Posted by Lee on 2003-12-22 06:54:26