I am struggling to understand how to sum over microstates in statistical mechanics.
Consider an $N$-spin system where $N \gg 1$ and $\Gamma=\{n_i \}$ for $1 \leq i \leq N$ and each $n_i$ is equal to $0$ or $1$. We want to compute the sum over the microstates $$\sum_{\Gamma}F(\Gamma)=\sum_{\{n_i\}}F(\{n_i\})$$ of a given function $F(\Gamma)$. Evaluate the sum over microstates for $F(\Gamma)=1$, $F(\{n_i\})=a{{\sum_{i=1}^N}n_i}, a>0$.
Question 1) I cannot see why we specify the functions on both sides of the summation, I would have thought specifying only would suffice.
Now it is clear to me that $\sum_{\Gamma}F(\Gamma)=\sum_{\Gamma}1=2^N$, but apparently,
$$\begin{align}\sum_{\{n_i\}}F(\{n_i\}) &= a{{\sum_{i=1}^N}n_i} \end{align}$$
so
$$\sum_{\Gamma} \left( a{\sum_{i=1}^N} n_i \right) = \sum_\Gamma a(n_1+n_2+ \dots n_N)=a\sum_{i=1}^N \sum_{\Gamma}n_i$$
This I am happy with, but apparently
$$\sum_{\Gamma}n_1=\sum_{n_1=0}^1 n_1 \sum_{n_2=0}^1 1 \dots \sum_{n_N=0}^1 1 = 2^{N-1} $$
and so on for $n_2,n_3,\dots$.
Question 2) I cannot see why we have fixed $n_2 \dots n_N$ to be equal to $1$?