Quick formula for $f(n) = \sum_{i=0}^{i=n}{k^{ji}}$ where $k \in \boldsymbol{\mathbb{R}}$ & $j \in \boldsymbol{\mathbb{N}}$

Question

I am trying to figure out a function $f(n)$ which takes the input $n$, where $n \in \boldsymbol{\mathbb{N}}$, and outputs the sum $\sum_{i=0}^{i=n}{k^{ji}}$ till the $n^{th}$ value where $k \in \boldsymbol{\mathbb{R}}$ & $j \in \boldsymbol{\mathbb{N}}$. For example $\sum_{i=0}^{i=n}{3^{2i}}$

Answer 1

Just figured it out myself. Let $a \in \boldsymbol{\mathbb{R}}$.

$$S(n) = ak^{0} + ak^{j} + ak^{2j} + ak^{3j} + \dots + ak^{jn}$$ $$k^{j}S(n) = ak^{j} + ak^{2j} + ak^{3j} + ak^{4j} + \dots + ak^{jn+j}$$ So, $$S(n) - k^{j}S(n) = ak^{0} - ak^{jn+j}$$ $$S(n)(1 - k^{j}) = ak^{0} - ak^{jn+j}$$ $$S(n) = \frac{ak^{0} - ak^{jn+j}}{(1 - k^{j})}$$ $$S(n) = \frac{a(1 - k^{jn+j})}{(1 - k^{j})}$$