Question

Find a closed-form formula that reproduces the sequence 1,-1, 1, -1, 1, -1, 1, ... (zero-based indexing). Check all that are correct.

Answer 1

Recall that

$\cos(0) = \cos(2\pi) = \cos(4\pi) = \cos(2n\pi) = 1$

and

$\cos(\pi) = \cos(3\pi) = \cos(5\pi) = \cos((2n+1)\pi) = -1$

Then the given sequence could be written as

$a_n = \cos(n\pi) = (-1)^n$