An algebra problem by Anthony Pham
There is a sequence of integers such that , for every . If the sum of the first 1996 terms is 2015, and the sum of the first 2015 terms is 1996, what is the sum of the first 2019 terms?
The answer is 38.
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1 solution
We note that the sequence is 6-periodic, in the form a 1 , a 2 , a 2 − a 1 , − a 1 , − a 2 , a 1 − a 2 , a 1 , a 2 , …
Therefore, the sum of the first n terms is simply the sum of the first n mod 6 terms.
Thus, we are given that the sum of the first 4 terms is 2 0 1 5 and the sum of the first 5 terms is 1 9 9 6 . That is, 2 a 2 − a 1 = 2 0 1 5 a 2 − a 1 = 1 9 9 6 . We solve and find a 2 = 1 9 . Finally, we are asked for the sum of the first 2 0 1 9 - i.e. the sum of the first 3 - terms. This is given by 2 a 2 = 3 8 .