Difference equation formula $\sum a^t = \frac{a^t}{a-1}$.

Question

As I explain below, this question was originally posted by user YYG, but then deleted. I am reposting the question (from memory) and I will answer it myself below.

Question: In Difference Equations by Walter Kelley and Allan Peterson, 2ed., p. 22, why is it that for $\Delta C(t) = 0$ and $a \ne 1$, we have

$$\sum a^t = \frac{a^t}{a - 1} + C(t)?$$

Note (providing context for why I am asking this question): User YYG, with whom I am not acquainted, asked two questions that he then deleted after receiving multiple downvotes, before I was able to answer the second.

The questions he asked were legitimate ones, but had the appearance of being nonsense if one was unfamiliar with the notations used in difference equations, as he himself explained. I find it regrettable that, in all likelihood, YYG withdrew the questions because of the comments and feedback he received, and that is why I have chosen to ask and answer this question myself.

Answer 1

Take $\Delta$ of the function appearing on the right-hand side and check that you get $a^t$.

$$\Delta \left( \frac{a^t}{a-1} + C(t) \right) = \frac{a^{t+1}}{a-1} - \frac{a^t}{a-1} + \Delta C(t) = \frac{a^{t+1} - a^t}{a-1} = \frac{a^t(a-1)}{a-1} = a^t.$$