This is related to this other question of mine Showing that $\operatorname {Br}(\Bbb F_q)=0$ in which I also got stuck at writing a free resolution. I want to compute the group cohomology $H^\bullet(\Bbb Z/n\Bbb Z) = H^\bullet(\Bbb Z/n\Bbb Z, \Bbb Z)$ and, more specifically, show that it is non zero in infinitely many levels.
I have $H^\bullet(\Bbb Z/n\Bbb Z) = \operatorname{Ext}_{\Bbb Z[\Bbb Z/n\Bbb Z]} (\Bbb Z, \Bbb Z)$. So I look for a free $\Bbb Z[\Bbb Z/n\Bbb Z]$-resolution of $\Bbb Z$. Let me write $C_n = \Bbb Z/n\Bbb Z$ for clarity and pick a generator $g\in C_n$.
$$...\to \oplus_{i=1}^n \Bbb Z[C_n] \xrightarrow{r_1} \Bbb Z[C_n] \xrightarrow{r_0} \Bbb Z \to 0$$
where $r_0$ sends $g\mapsto 1$ and $r_1$ sends the generator $g_i$ of the $i$-th summand to $g^i -1$. But I am stuck in trying to write the next term (and hopefully see a periodicity).
There is a 4-term exact sequence $$ 0\to\mathbb Z\to\mathbb Z[C_n]\xrightarrow{1-\sigma}\mathbb Z[C_n]\to\mathbb Z\to0 $$ giving rise to a 2-periodic free resolution of $\mathbb Z$.