klastiesym

## Answered question

2022-10-19

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

### Answer & Explanation

Jimena Torres

Beginner2022-10-20Added 20 answers

Let $a\in \mathbb{R}$.
$S\left(n\right)=a{k}^{0}+a{k}^{j}+a{k}^{2j}+a{k}^{3j}+\cdots +a{k}^{jn}$
${k}^{j}S\left(n\right)=a{k}^{j}+a{k}^{2j}+a{k}^{3j}+a{k}^{4j}+\cdots +a{k}^{jn+j}$
So,
$S\left(n\right)-{k}^{j}S\left(n\right)=a{k}^{0}-a{k}^{jn+j}$
$S\left(n\right)\left(1-{k}^{j}\right)=a{k}^{0}-a{k}^{jn+j}$
$S\left(n\right)=\frac{a{k}^{0}-a{k}^{jn+j}}{\left(1-{k}^{j}\right)}$
$S\left(n\right)=\frac{a\left(1-{k}^{jn+j}\right)}{\left(1-{k}^{j}\right)}$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?