Combi?
0 ≤ k , n ≤ 2 0 1 5 ∑ cos ( 2 0 1 6 2 π n k ) − sin ( 2 0 1 6 2 π n k )
The answer is 2016.
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1 solution
We know that the sum ∑ k = 0 2 0 1 5 ( e 2 π k i / 2 0 1 6 ) n of the n th powers of the 2016th roots of unity is 0 for 0 < n < 2 0 1 6 ; taking real and imaginary parts, we see that the given sum is 0 for n > 0 . For n = 0 the sum is 2 0 1 6 cos ( 0 ) − 2 0 1 6 sin ( 0 ) = 2 0 1 6
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Ya gotta admit that, so far, this was the shortest solution. Otto's is the next shortest.
Next year is 2016. Ergo, that's the answer.