Vardan95
In the sequence \(a_1 = 2017\) and \(a_n=\frac{a_{n-1} - 1}{a_{n-1}}\), what is the value of \(a_{2017}\)?
(A) \(-2017\)
(B) \(\frac{-1}{2017}\)
(C) \(\frac{2016}{2017}\)
(D) 1
(E) 2017
Dear
Vardan95,
I'm happy to respond.

This is a diabolically hard question, extremely clever! I don't know if this would appear on the GMAT--it really would be at about the outer level of what the GMAT would ask.
Here's my solution.
\(a_1 = 2017\)
\(a_2 = \dfrac{a_1 - 1}{a_1} = \dfrac{2016}{2017}\)
\(a_3= \dfrac{a_2 - 1}{a_2} = \dfrac{-1/2017}{2016/2017} = \dfrac{-1}{2016}\)
\(a_4 = \dfrac{a_3 - 1}{a_3} = \dfrac{-2017/2016}{-1/2016} = 2017\)
BINGO! That's the key to the problem! The sequence is cyclical, repeating with a cycle of three. For every n that is a multiple of 3, \(a_n = \dfrac{-1}{2016}\) and the following term will be \(a_{n+1} = 2017\).
Well, 2016 is a multiple of three, because 2 + 0 + 1 + 6 = 9, a multiple of three. Thus we know
\(a_{2016} = \dfrac{-1}{2016}\)
\(a_{2017} = 2017\)
Answer =
(E) A fantastic problem! Does this make sense?
Mike