my book gives me the following equation : $$ \sum_{i=0}^4 \frac{1}{3^i} = \sum_{i=0}^4 \left(\frac{1}{3} \right)^i. $$ And I do not know based on which rules they did that. I could see that because $1^i =1$ no matter what number $i$ is, the writers of the book wrote it like that. But what if the number in the numerator was another number lets say $2$? I mean would it still be allowed then?
Thanks in advance.
On
As you observed, you can rewrite ${1\over3^i}$ as $\left(1\over3\right)^i$ because $1^i=1$ for all $i$, and as pointed out in comments, you cannot do the same for something like $2\over3^i$ because $2^i\not=2$ in general.
It might help to discuss why the book bothered to rewrite ${1\over3^i}$ as $\left(1\over3\right)^i$. Most likely, the reason is to apply the formula for geometric progressions,
$$\sum_{i=0}^Nx^i={1-x^{N+1}\over1-x}$$
with $x={1\over3}$ (and $N=4$), so that
$$\sum_{i=0}^4{1\over3^i}=\sum_{i=0}^4\left(1\over3\right)^i={1-(1/3)^5\over1-(1/3)}={243-1\over243-81}={242\over162}={121\over81}$$
Now what if there is a $2$ instead of a $1$ in the numerator? In that case, the first step is to move the $2$ outside the summation, i.e.,
$$\sum_{i=0}^4{2\over3^i}=2\sum_{i=0}^4{1\over3^i}=2\sum_{i=0}^4\left(1\over3\right)^i=2\cdot{1-(1/3)^5\over1-(1/3)}=\cdots={242\over81}$$
Simply note that for any $n\ge 0$ we have $1^n=1$ therefore
$$\frac{1}{3^{n}}=\frac{1^n}{3^{n}}=\left(\frac{1}{3}\right)^{n}$$
and more in general for $b\neq 0$
$$\frac{a^n}{b^{n}}=\left(\frac{a}{b}\right)^{n}$$
indeed for $n$ integer by definition of exponentiation
$$\frac{a^n}{b^{n}}=\frac{\overbrace{a\cdot a\cdot\ldots\cdot a}^{\color{blue}{\text{n terms}}}}{b\cdot b\cdot\ldots\cdot b}=\overbrace{\frac{a}{b}\cdot \frac{a}{b}\cdot\ldots\cdot \frac{a}{b}}^{\color{blue}{\text{n terms}}}=\left(\frac{a}{b}\right)^{n}$$