Andre Alexander Bell, 18.06.2010 11:23:
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...
The thing is, if you can't do the math in the time that your processor
needs to run the brute force loop, it's often not worth doing the math at all.
Stefan
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