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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 ?