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Old 2008-03-15, 07:35   #4
nibble4bits
 
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Nov 2005

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Do I understand right? That this is based on this formula for quadratics:
(n+1)^2=n^2+2n+1; let m=n+1
m^2-n^2-1=2n
Thus consecutive squares are always 2n+1 appart.

That formule you mentioned is equivilent to this code:
Code:
Let x = 0
For i = 1 to n
x = x + i + i - 1
Next i
REM Returns in x, the square of n using recursive addition instead of multiplication.
The optimizes trivially without multiplication to:
Code:
Let x=0
For i = 1 to n
x = x + i + i
Next i
x = x - n
You can further remove the For...Next loop using an identity but you get this:
Let x = 2 * n * (n + 1) / 2 - n
Or:
Code:
Let x = n * (n + 1) - n
Or:
Code:
Let x = n^2 + n - n
As you can see, this leads you back to your original equation of x=n^2.



EDIT:
You should note that for m=n+1:
m^3 - n^3 - 1 =
(n+1)^3 - n^3 - 1 =
n^3 + 3n^2 + 3n + 1 - n^3 - 1 =
3(n^2 + n) =
3n(n+1)

Last fiddled with by nibble4bits on 2008-03-15 at 07:44 Reason: Added note
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