General formula for nth element of the sequence 0, 1, 0, 1, ... The sequence is $f = 0, 1, 0, 1, \ldots$
I want to find a general formula for the $n$th element. The sequence starts at $n = 0$ (the $0$ here is not the first element $0$ but rather denotes the $0$th position).
One easy and obvious solution is: $n$th $f = n \bmod 2$. This works because even positions have $0$ and odd positions have $1$.
However, this question is part of a homework and modulus has not been discussed (or part of the syllabus or even a prerequisite). And so I am hesitant to use it.
Is there another way to solve this problem using only basic arithmetic operations (one that a beginning high schooler knows of)?

You could compact the alternating sign formula with  $$f(n) = \sin \left( \frac{n \pi}{2} \right)^2$$
Applying $$(\sin(x))^2= \left( \frac{1}{2} \right)(1 - \cos(2x)) $$
So you get $0, 1, 0, 1, 0, 1, \ldots$
