Sequence

Hey,guys i once saw this problem at my national math olympiad ans wanted to know if it is stated right: Consider the infinite sequence defined as : $a_1=2$,$a_2=3$ and $a_{k+2}=\frac{a_{k+1}}{a_k}$ for $k >= 3$.Find $a_{2015}$.Is it solvable?

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Yes, it is solvable. Also, Hint:- Try solving for a few terms.

Siddhartha Srivastava - 6 years, 4 months ago

Proceed,please!

lawrence Bush - 6 years, 3 months ago

$a_1 = 2, a_2 = 3, a_3 = \frac{3}{2}, a_4 = \frac{1}{2}, a_5 = \frac{1}{3}, a_6 = \frac{2}{3}, a_7 = 2, a_8 = 3$

So the sequence is cyclic every $6$ terms.

Therefore, $a_{2015} = a_5 = \frac{1}{3}$

Siddhartha Srivastava - 6 years, 3 months ago

now i know

NJ Ibera - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$