gr.group theory - (co)homology of cyclic groups

The answer to your question is simple. Choose a generator $sigma$ of $G$. Then we have an exact sequence $$0 to mathbb{Z} xrightarrow{m mapsto m 1_G} mathbb{Z}[G] xrightarrow{sigma-1} mathbb{Z}[G] xrightarrow{g in G mapsto 1} mathbb{Z} to 0.$$ Since $mathbb{Z}[G]$ is a free (hence projective) $mathbb{Z}[G]$ module, this means that we have isomorphisms $mathrm{Ext}^n(mathbb{Z},M) cong mathrm{Ext}^{n+1}(mathrm{Im}(sigma-1),M) cong mathrm{Ext}^{n+2}(mathbb{Z},M)$ for $n ge 1$, and since $mathrm{Ext}(mathbb{Z},M) cong H^n(M)$, this gives us the desired periodicity.

This is related to the bar resolution in the sense that the bar resolution gives us group cohomology specifically because $mathrm{Ext}^n(mathbb{Z},M) cong H^n(G,M)$. It follows that $mathrm{Ext}$ can be computed by applying $mathrm{Hom}(-,M)$ to a projective resolution of $mathbb{Z}$, and the bar resolution is precisely such a resolution.

Note that by $mathrm{Ext}$ we mean over the category $mathbb{Z}[G]$-Mod.