$\sum_\eta \exp(b*(\eta_1 + ...+\eta_n)) = (1+e^b)^n$.

Question

I'm wondering why this is a correct relation, $\eta$ is the collection of $\{\eta_1,...,\eta_n \}$; ${\eta_1,...,\eta_n }$ are independent and the given relation is : $\sum_\eta \exp(b*(\eta_1 + ...+\eta_n))= \sum_{\eta_1} \sum_{\eta_2} ...\sum_{\eta_n} \exp(b*(\eta_1 + ...+\eta_n)) = (1+e^b)^n$. Every $\eta_i = \{1,0\}$

Thanks in advance

Answer 1

At present what you have is unlikely to be true. To test this out you can look at the case of $n=1$ then your sum is $$\sum_{\eta_1} e^{b \eta_1} = e^b+e^{-b} \neq e^{b}+1$$

Possibly your $\eta$ need to be from $\{0,1\}$. Or there is some other relationship/constraint on the $\eta$