I'm working with the inhomogeneous bar resolution, defined as $\cdots \to L_n \overset{\partial_n}\to L_{n-1} \to \cdots \to L_0 \overset{\varepsilon} \to R \to 0$ where $L_i$ is the free $RG$-module over $i$ copies of $G$, and in particular $L_0$ is a free module over the empty symbol $()$. Here, $$\partial_n ((x_1 , \ldots, x_n)) = x_1 (x_2 , \ldots, x_n) + \sum_{i=1}^{n-1} (-1)^i (x_1 , \ldots, x_i x_{i+1} , \ldots, x_n) + (-1)^n (x_1 , \ldots, x_{n-1})$$
and $\varepsilon(1()) = 1$. I'm trying to show that the collection of maps $\gamma_n(x_0 (x_1 , \ldots, x_n) = (x_0 , x_1 , \ldots, x_n), \gamma_{-1}(x) = x()$ is a contracting homotopy, and I'm mostly done, except for the case $\partial_1 \gamma_0 + \gamma_{-1} \varepsilon = id$. My problem is that I'm not entirely sure how to interpret $\partial_1$ by the definition above. It seems like it should be $\partial_1 ((x_1)) = x_1 () + (-1)^1 ()$ since $n-1 = 0$ makes the sum in the middle disappear, but using this interpretation with the contracting map $\gamma_0(x_0 ()) = (x_0)$ doesn't work. The interpretation that it should just be $\partial_1((x_1)) = x_1()$ also doesn't seem to work, however. Should I be interpreting the sum in the middle for the $n=1$ case some other way, or am I doing something else wrong?
Your first interpretation is correct: $\partial_1((x_1))=x_1()-()$. I think you may be misinterpreting the definition of $\varepsilon$: note that $\varepsilon$ is supposed to be a map of $RG$-modules, so $$\varepsilon(x())=x\cdot\varepsilon(())=x\cdot 1=1$$ for any $x\in G$, since $G$ acts trivially on $R$. We thus have $$\partial_1\gamma_0(x_0())+\gamma_{-1}\varepsilon(x_0())=\partial_1((x_0))+\gamma_{-1}(1)=x_0()-()+()=x_0()$$ for any $x_0\in G$, and so $\partial_1\gamma_0+\gamma_{-1}\varepsilon$ is the identity since it is $R$-linear.