Let $G = \langle g \rangle$ be a cyclic group of order $n$. Consider the free ring $\mathbb Z[G]$ of all formal sums of elements from $G$ with coefficients from $\mathbb Z$, with multiplication given by lineary extending the multiplication given in the group $G$.
Let $\varepsilon : \mathbb Z[G] \to \mathbb Z$ be given by $$ \varepsilon \left(\sum_{i=0}^{n-1} a_i g^i\right) := \sum_{i=0}^{n-1} a_i $$ and $\psi : \mathbb Z[G] \to \mathbb Z[G]$ by $$ \psi\left(\sum_{i=0}^{n-1} a_i g^i\right) = \left(\sum_{i=0}^{n-1} a_i g^i\right) \cdot \left( g - 1 \right) = \sum_{i=0}^{n-1} (a_{i-1} - a_i)g^i $$ with $a_{-1} := a_{n-1}$. I am reading a proof where these maps occur, and the author wants to show that $\mbox{ker}(\varepsilon) = \mbox{im}(\psi)$. It goes like this \begin{align*} \mbox{ker}(\varepsilon) & = \left\{ \sum_{i=0}^{n-1} a_i g^i : \sum_{i=0}^n a_i = 0 \right\} \\ & = \left\{ (a_0 + (a_0 + a_1)g + \ldots + (a_0 + \ldots + a_{n-1})g^{n-1})(g - 1) \mid a_i \in \mathbb Z \right\} \\ & = \mbox{im}(\psi). \end{align*} Why he writes this particular expression $$ a_0 + (a_0 + a_1)g + \ldots + (a_0 + \ldots + a_{n-1})g^{n-1}? $$ I see that if we denote by $\sum_{i=0}^{n-1} b_i g^i$ the image under $\psi$, then \begin{align*} b_0 & = (a_0 + \ldots + a_{n-1}) - a_0 = a_1 + \ldots + a_{n-1} \\ b_1 & = a_0 - (a_0 + a_1) = -a_1 \\ b_2 & = (a_0 + a_1) - (a_0 + a_1 + a_2) = -a_2 \\ \vdots \\ b_{n-1} & = (a_0 + \ldots a_{n-2}) - (a_0 + \ldots + a_{n-1}) = -a_{n-1} \end{align*} hence $b_0 + b_1 + \ldots + b_{n-1} = 0$, but this holds for every image, i.e. as $\psi\left(\sum_{i=0}^{n-1} a_i g^i\right) = \sum_{i=0}^{n-1} (a_{i-1} - a_i)g^i$ we see easily that the coefficients of the image add to zero. Conversely, if we have some sum $\sum_{i=0}^{n-1} a_i g^i$ whose coefficients add to zero, then we could choose any sequence $b_1, \ldots, b_{n-1}$ of coefficients with $a_i = b_{i-1} - b_i$ ($b_{-1} := b_{n-1}$), and these equations give a certain degree of freedom, i.e. there are many possible choices of coefficients possible, but the choise made above, i.e. $b_0 = a_0, b_1 = a_0 + a_1, \ldots, b_{n-1} = a_0 + \ldots + a_{n-1}$ does not give the original coefficients, so to me this choice looks arbitrary, I do not see the reason to write it here.
I see that the equality is easy to see, by the arguments I wrote above, but I am interested what might have the author in mind as he wrote these coefficients appearing in the expression? Why this expression. Did it help to see anything I do not see?
Maybe you can better see this with matrices. Considering each element of $\mathbb{Z}[G]$ as a column vector (the rows correspond to the powers of $g$), the matrix associated to $\varepsilon$ is simply $A=\begin{bmatrix}1 & 1 & \dots & 1\end{bmatrix}$, whereas the matrix associated to $\psi$ is $$ B=\begin{bmatrix} -1 & 0 & \dots & 0 & 1\\ 1 & -1 & \dots & 0 & 0 \\ 0 & 1 & \dots & 0 & 0 \\ 0 & 0 & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & -1 & 0 \\ 0 & 0 & \dots & 1 & -1 \end{bmatrix} $$ Clearly, all columns of $B$ belong to the null space of $A$. If you sum the first row to the second row, the (new) second row to the third row, and so on, you get the matrix $$ \begin{bmatrix} -1 & 0 & \dots & 0 & 1\\ 0 & -1 & \dots & 0 & 1 \\ 0 & 0 & \dots & 0 & 1 \\ 0 & 0 & \dots & 0 & 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & -1 & 1 \\ 0 & 0 & \dots & 0 & 0 \end{bmatrix} $$ which clearly has rank $n-1$.
This proves the equality as well as the proof with elements.
For the other proof, consider $t=\sum_{i=0}^{n-1}a_ig^i\in\ker\varepsilon$; you want to find $u=\sum_{i=0}^{n-1}b_ig^i$ such that $u(g-1)=t$; this is equivalent to asking that \begin{cases} b_{n-1}-b_0=a_0\\ b_0-b_1=a_1\\ b_1-b_2=a_2\\ \vdots\\ b_{n-2}-b_{n-1}=a_{n-1} \end{cases} and you should clearly see the same matrix as above.