If you subtract the sum of the first x numbers, squared, from the sum of the next x numbers, squared, (for example, 6² + 5² + 4² - 3² - 2² - 1²); you will get (x²)(2x+1) and also x²(x+1)² minus x^4. Prove why this works.

We're evaluating Sum = (2x)²+(2x-1)²+...+(x+1)²-x²-(x-1)²-...-1²
Grouping up the terms: Sum = [(2x)²-x²]+[(2x-1)²-(x-1)²]+...+[(x+1)²-1²]
Each bracket [] contains a difference of squares, so: Sum = (2x+x)*(2x-x)+(2x-1+x-1)*(2x-1-x+1)+...+(x+1+1)*(x+1-1) = 3x*x+(3x-2)*x+...+(x+2)*x = x*[(x+2)+(x+4)+...+(3x-2)+3x]
Re-grouping inside the brackets: Sum = x*[(x+x+...+x)+2+4+...+(2x-2)+2x] = x*[(x+x+...+x)+2(1+2+...+(x-1)+x)]
There are x x's in the first parentheses, and the sum inside the second parentheses is x*(x+1)/2. Thus: Sum = x*[x²+x(x+1)] = x²*(x+x+1) = x²(2x+1)
For the second result: x²(x+1)²-x^4 = x²(x²+2x+1)-x^4 = x^4+2x³+x²-x^4 = 2x³+x² = x²(2x+1)
And that about does it :)

Posted by zaphod on 2003-12-24 01:51:43