On 18/06/2010 10:23, Andre Alexander Bell wrote:
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
The good news is that this is easily the fastest piece of code that I've
seen yet. The bad news is that first prize in the speed competition is
a night out with me. :)
Cheers.
Mark Lawrence.
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