I'm following Evens' The Cohomology of Groups, section 2.1 where resolutions for finite cyclic groups are developed. I understood the final result for a finite cyclic group $G$ and a $G$-module $M$:
$$H^n(G, M) = \begin{cases} M^G\text{, if }n = 0\\ M^G/TM \text{, if }n\text{ is even and non-zero}\\ _TM/IM\text{, if }n\text{ is odd}\end{cases}$$
Where $M^G$ represents the submodule of fixed points, $TM$ is the image of the action restricted to the group ring ideal $T = \{\sum_{1}^{|G|} g^i| G = \langle g \rangle\}$, $IM$ is the image of the action restricted to the augmentation ideal $I$ and $_TM$ is the submodule $\{m \in M| Tm = 0\}$. An exercise then follows:
Show that if $G$ is a finite cyclic group and $M$ is a $G$-module, then $|G|H^n(G, M) = 0$ for $n > 0$. Hint: for $n$ odd, consider $|G|x - Tx$ for $x \in M$.
I've already solved it for $n$ even. It took me awhile to snap it, but it was pretty straightfoward once I've got a grip of what was going on. A week rolled over and I still can't figure out the odd case. Any hints?
Remember that the augmentation ideal $I$ is generated over $\mathbb{Z}$ by the elements of the form $1 - g$ for $g\in G$. So if you take an element $x\in M$ such that $T (x) = 0$, then $$|G|\cdot x = |G|\cdot x - T (x) = |G|\cdot x - \sum_{g\in G} g\cdot x = \sum_{g\in G} (1 - g)\cdot x \in IM.$$ (Sorry for not following your notation here, my $g$ is an arbitrary element of $G$, not a generator.)
By the way, it is true for any finite group that multiplication by $|G|$ kills $H^n (G,M)$ and $H_n (G,M)$ for $n > 0$. You will probably see a proof of this later (you can show this using the bar-resolution, as in [Weibel, An Introduction to Homological Algebra, Theorem 6.5.8], or using transfer maps, as in [Weibel, An Introduction to Homological Algebra, Exercise 6.7.8]).