On 06/16/2010 12:47 PM, Lie Ryan wrote:
> Probably bending the rules a little bit:
>
>>>> sum(x**2 - 8*x - 20 for x in range(1, 2010, 5))
> 536926141
Bending them even further, the sum of the squares from 1 to N is given by
(1) N*(N+1)*(2*N+1)/6.
The given problem can be divided into five sums of every fifth square
starting from 1,2,3,4,5 respectively. This is given by
(2) S(a,k) = sum_{i=1}^k (5*i-4+a)^2
for a in range(5) and we are finally interested in
(3) S(k) = S(0,k) + S(1,k) + S(2,k) - S(3,k) - S(4,k)
Substituting (2) and (1) in (3) gives (in python code)
>>> S = lambda k: (50*k**3 - 165*k**2 - 47*k) / 6
>>> S(2010/5)
536926141
However, this only works for full cycles of (1,1,1,-1,-1) and you would
have to add/subtract the modulus parts yourself. (e.g. if you are
interested in your sum from 1..2011...
Cheers
Andre
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