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The sum of 2016 terms (Posted on 2017-03-21) Difficulty: 3 of 5
The sequence a1, a2, a3 ..., of integers has the property
that for all n ≥ 3, an=an-1-an-2.
If the sum of the first 1807 terms is 1807 and the sum of the first 1907 terms is 1907, what is the sum of the first 2016 terms?

No Solution Yet Submitted by Ady TZIDON    
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Solution Solution Comment 1 of 1
The sequence repeats the six values a1, a2, a2-a1, -a1, -a2, a1-a2

The sum of these six terms is zero.  2016 mod 6 = 0, therefore the sum of the first 2016 terms is 0.

Going further, the sequence of partial sums repeats the six values a1, a1+a2, 2*a2, 2*a2-a1, a2-a1, 0.  1807 mod 6 = 1 and 1907 mod 6 = 5.  Then a1=1807 and a2-a1 = 1907, which implies a2 = 3714.

The sequence repeats 1807, 3714, 1907, -1807, -3714, -1907 and the sequence of partial sums repeats 1807, 5521, 3814, 2007, 1907, 0.

  Posted by Brian Smith on 2017-03-21 09:42:36
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