One-dimensional with lattice constant (Problem)

[Another]

In question (a) Calculate the average position ?
https://en.wikipedia.org/wiki/Expected_value#Finite_case
$<x> =\sum x_ip_i$
I saw the answer and wondered.
$\bar{x} = <x> = \sum_{n=0}^{N} \frac{N!}{n!(N-n)!} x_n p^n q^{N-n}$
so $x_n = (2n-N)a$ a is lattice constant

why x_n = (2n-N)a ? why is that ?

 

[Lord Jestocost]

When after N steps of length a there were n steps to the right and (N-n) steps to the left, the distance x_n traveled by the atom (starting at the origin x = 0) will be
x_n = (na – (N-n)a) = (2n-N)a.