I'm trying to solve an exercise in Morandi's "Field and Galois Theory", which guides one through the proof that the second cohomology group of $G=\langle\sigma\rangle$ finite is $A^G/NA$, where $A^G=\{\sigma a=a\}$ and $N=\sum_{j=0}^{n-1}\sigma^j\in\mathbb Z[G]$. Morandi does this the following way
Show that $A^G\to H^2(G,A),\,\,a\mapsto [f_a]$, where $$f_a(\sigma^i,\sigma^j)=\begin{cases}0,&0\leq i+j<n\\a,&n\leq i+j<2n-1\end{cases}$$ is a group homomorphism. Show that $NA$ is its kernel. Show that if $f\in Z^2(G,A)$ and $a=\sum_{j=0}^{n-1}f(\sigma^j,\sigma)$, then $[f]=[f_a]$.
I've showed the first two statements and that $\sum_{j=0}^{n-1}f(\sigma^j,\sigma)\in A^G$, however I'm having trouble showing $f\sim f_a$. Morandi provides the following hint
Make use of the cocyle condition $$\sigma^if(\sigma^j,\sigma)-f(\sigma^{i+j},\sigma)+f(\sigma^i,\sigma^{j+1})-f(\sigma^i,\sigma^j)=0.$$
I suppose the idea is to sum these over all $i,j$, then break the sums apart depending on whether $i+j<n$ or $i+j\geq n$, but I haven't been able to make this work yet.
Note that I am familiar with the resolution argument (any projective resolution yields the same cohomology groups, use a specific resolution for the case of $G$ cyclic) and am looking strictly for an elementary proof of the above.