Compute base $q$ expansion of $\frac 12$, where $q=2n+1$

Question

Problem Verbatim:

Let $q=2n+1$ for some natural number $n$. Compute the base $q$ expansion of $\dfrac{1}{2}$.

In the answer key the only information it gives is

if $q=1\pmod k$, then the base $q$ expansion of $\dfrac{1}{k}$ is $\dfrac{q-1}{k}$ in each position.

However, it does not give any information as to how they came to that answer. I tried to work it out backwards for myself, but the furthest I can get is using the definition of congruence mod $k$ to get $n=\frac{q-1}{k}$. So why is every position in the expansion equal to $n$?

Any help as to how they came to this conclusion would be appreciated.

Answer 1

Hint: $$\sum_{i=1}^{\infty}q^{-i} = \frac{q^{-1}}{1-q^{-1}}=\frac{1}{q-1} \implies \frac{q-1}{k} \sum_{i=1}^{\infty}q^{-i} = \frac{1}{k}.$$