klastiesym
2022-10-19
Answered

Try to figure out a function $f(n)$ 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 $\sum _{i=0}^{i=n}{3}^{2i}$

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Jimena Torres

Answered 2022-10-20
Author has **20** answers

Let $a\in \mathbb{R}$.

$$S(n)=a{k}^{0}+a{k}^{j}+a{k}^{2j}+a{k}^{3j}+\cdots +a{k}^{jn}$$

$${k}^{j}S(n)=a{k}^{j}+a{k}^{2j}+a{k}^{3j}+a{k}^{4j}+\cdots +a{k}^{jn+j}$$

So,

$$S(n)-{k}^{j}S(n)=a{k}^{0}-a{k}^{jn+j}$$

$$S(n)(1-{k}^{j})=a{k}^{0}-a{k}^{jn+j}$$

$$S(n)=\frac{a{k}^{0}-a{k}^{jn+j}}{(1-{k}^{j})}$$

$$S(n)=\frac{a(1-{k}^{jn+j})}{(1-{k}^{j})}$$

$$S(n)=a{k}^{0}+a{k}^{j}+a{k}^{2j}+a{k}^{3j}+\cdots +a{k}^{jn}$$

$${k}^{j}S(n)=a{k}^{j}+a{k}^{2j}+a{k}^{3j}+a{k}^{4j}+\cdots +a{k}^{jn+j}$$

So,

$$S(n)-{k}^{j}S(n)=a{k}^{0}-a{k}^{jn+j}$$

$$S(n)(1-{k}^{j})=a{k}^{0}-a{k}^{jn+j}$$

$$S(n)=\frac{a{k}^{0}-a{k}^{jn+j}}{(1-{k}^{j})}$$

$$S(n)=\frac{a(1-{k}^{jn+j})}{(1-{k}^{j})}$$

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