If a_(n+1) = a_(n) / a_(n-1), find a_2017

We are given; a₁ = 2, a₂ = 6 a_n+1 = a_n / a_n-1 And we want to find a₂₀₁₇. The question is hard because if we tried simply applying the formula to get an answer, we would have to do it thousands of times. We want a_n+1 to depend on other terms of the sequence as little as possible. Right now it depends on two of them. But if we apply the recursive formula to a_n, we get this: a_n+1 = a_n / a_n-1 = (a_n-1 / a_n-2) / a_n-1 = 1/a_n-2         (only works if n ≥ 3) Equivalently: a_n+3 = a_n            (for all n ≥ 1) This is perfect! It means that if you only look at every third term of the sequence, the terms just keep being "flipped". a₁ = 2 a₃₊₁ = 1/2 a₆₊₁ = 2 a₉₊₁ = 1/2 ... a₂₀₁₆₊₁ = 2 This works out because 2016 is an even multiple of 3, so we know that the 2 has been inverted an even amount of times - that's the same as not flipping it at all. Awesome right?