Sequence for 2017!

Algebra Level 4

A sequence is determined by n 1 = 2017 n_1=2017 and n k + 1 = 1 + n k 1 − n k n_{k+1}=\dfrac{1+n_{k}}{1-n_{k}} for k ≥ 1 k \geq 1 .

What is the value of n 2017 n_{2017} ?


The answer is 2017.

View wiki
Freddie Hand by Freddie Hand , 18, United Kingdom

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Freddie Hand
Jan 23, 2017

We find the answer by re-iterating the sequence.

n k + 2 = 1 + 1 + n k 1 − n k 1 − 1 + n k 1 − n k = 2 − 2 n k = − 1 n k n_{k+2}=\frac{1+\frac{1+n_{k}}{1-n_{k}}}{1-\frac{1+n_{k}}{1-n_{k}}}\\=\frac{2}{-2n_{k}}\\=-\frac{1}{n_{k}}

Also, n k + 4 = − 1 − 1 n k = n k n_{k+4}=-\frac{1}{-\frac{1}{n_{k}}}=n_{k}

Therefore, this sequence has a period of 4.

As 2017 ≡ 1 ( m o d 4 ) 2017 \equiv 1 \pmod{4} , n 2017 = n 1 = 2017 n_{2017}=n_{1}=2017

4 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...