TITLE: Proof by induction on Fibonacci numbers: show that $f_n\mid f_{2n}$
QUESTION [12 upvotes]: I was studying Mathematical Induction when I came across the following problem:

The Fibonacci numbers are the sequence of numbers defined by the linear recurrence equation-
$f_n = f_{n-1} + f_{n-2} $ with $f_1 = f_2 = 1$
Use induction to show that $f_n \; | \; f_{2n}$ ($f_n$ divides $f_{2n}$)

Basis Step is obviously true; but I'm facing difficulty in the Inductive Step. If I assume the inductive hypothesis to be true for some $k$, i.e., $ \dfrac{ f_{2k} } { f_{k} } = c$ (For some positive integer $c > 0$), I'm not clear as to how I should proceed further and prove that $P(2k+1)$ is also true.
I'm new here, so if I'm doing anything wrong, please overlook it on the account of my naivety.

REPLY [7 votes]: The question is old, Calvin Lin's answer is great and already accepted but here is another method (for the famous sake of completess): 
We know that $f_n \wedge  f_m=f_{n\wedge m}$, where $a\wedge b$ is the gcd of $a$ and $b$.
So $f_n \wedge f_{2n}=f_{n\ \wedge\ 2n}=f_n$. This means that $f_n$ divides $f_{2n}$.