Prove that $2^n+(-1)^{n+1}$ is divisible by 3. Prove that $2^n+(-1)^{n+1}$ is divisible by 3 for $n\in\mathbb{N}$.
My attempt:
For $n=1$:
$2^1+(-1)^2 = 2 + 1 = 3, 3 |3$
We assume that $3|(2^n+(-1)^{n+1})$
Then for $n+1$:
$2^{n+1} + (-1)^{n+2} = 2\cdot 2^n -(-1)^{n+1}$
I am trying to show that for $n+1$, the expression is a constant multiple of the case for $n$ plus or minus a multiple of $3$. However, the minus sine I pulled out of the alternating term is tripping me up. Am I going about this the right way?

If $n=1$, then $2^{n} + (-1)^{n+1} = 3$, which is divisible by $3$; if $n \geq 1$ such that $2^{n} + (-1)^{n+1} = 3q$ for some integer $q \geq 1$, then
$$
2^{n+1} + (-1)^{n+2} = 2[ 3q - (-1)^{n+1} ] + (-1)^{n+2} = 6q + 2(-1)^{n+2} + (-1)^{n+2} = 6q + 3(-1)^{n+2},
$$
which is again divislble by $3$.
