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In the sequence a_1=2017 and . . .

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In the sequence a_1=2017 and . . .

by M7MBA » Sun Oct 29, 2017 7:24 am
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) −1/2017

(C) 2016/2017

(D) 1

(E) 2017

The OA is E.

Experts may you give me some help here. I don't know why the correct option is E?
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by ErikaPrepScholar » Mon Oct 30, 2017 7:39 am
Hey M7MBA,

When you have a function that you can't seem to simplify, and you're asked to find a crazy high value, the usual trick is to find a pattern.

Here, we have a_1, so let's find a_2:

$$a_2 = \frac{2017\ -\ 1}{2017} = \frac{2016}{2017}$$

No pattern yet, so we'll keep going:

$$a_3 = \frac{\frac{2016}{2017}\ -\ 1}{\frac{2016}{2017}} = \frac{\frac{-1}{2016}}{\frac{2016}{2017}} = \frac{-1}{2016}$$

And one more:

$$a_4 = \frac{\frac{-1}{2016}\ -\ 1}{\frac{-1}{2016}} = \frac{\frac{-2017}{2016}}{\frac{-1}{2016}} = 2017$$

So we see that a_4 = a_1. This means that the pattern will repeat again until a_7 = 2017, then a_10, and so on. So every time we add one to a multiple of three, we get 2017. Following that pattern on up, we see that a_2017=2017.
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