I'm doing a high school level question and one of the questions is figuring out which of the Fibonacci sequence numbers must be even. The 61st, 62nd, 63rd, 64th, or 65th. And the only information I'm given is how the sequence is defined recursively.
Apart from using my calculator and testing out each number, I'm not sure how I'm supposed to solve this without using college-level mathematics. (Btw, I'm allowed a calculator so maybe the question is just testing your ability to use it)
Since $F_{n+3}=2F_{n+1}+F_n$, it leaves the same remainder as $F_n$ on division by $2$. So the 61st term has the same parity as the 1st etc. (because $3|60$).
Let $F(n)$ denote the $n$th Fibonacci number. Note that: $$F(2)=1; F(3)=2$$ Then, note that an even number plus an odd number is odd, and an odd number plus an odd number is even. Shown by: $$[2\alpha+1]+2\beta=2(\alpha+\beta)+1 \text{ (odd)}$$ $$[2\alpha+1]+[2\beta+1]=2(\alpha+\beta+1)\text{ (even)}$$
Thus, the sequence alternates: $$E,O,O,E,O,O,E,O,O...$$ ($E$=Even, $O$=Odd)
The even numbers therefore are $F(3n), n\in\Bbb Z$, of which the ony one in your choices is $F(63)$ where $n=21$
$f_1=f_2=1$ and $$ f_n=f_{n-1}+f_{n-2} $$ for all $n > 2$
Note that an odd number plus an even number is odd and an odd plus an odd number is even.
So think of $a_1=a_2=\text{even}$ then $$ a_n=a_{n-1}+a_{n-2} $$
So you have $a_3=\text{odd}+\text{odd}=\text{even}$ and hence
$a_4=\text{even}+\text{odd}=\text{odd}$ and
$a_5=\text{even}+\text{odd}=\text{odd}$
Then it starts back over at $a_6=\text{odd}+\text{odd}$. So every multiple of $3$ will have $a_{3k}$ even. You can of course prove it, but I'm not entirely sure what constitutes high school math for you :)
