Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$

I'm an eight-grader and I need help to answer this math problem.

Problem:

Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$

This one is very hard for me. It seems unsolvable. How to calculate the series without using Wolfram Alpha? Please help me. Grazie!

Hint :

Let $$ S=\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots\tag1 $$ Dividing $(1)$ by $5^2$, we obtain $$ \frac{S}{5^2}=\frac{1}{5^3}+\frac{3}{5^5}+\frac{5}{5^7}+\frac{7}{5^9}+\frac{9}{5^{11}}+\cdots\tag2 $$ Subtracting $(2)$ from $(1)$, we obtain $$ S-\frac{S}{5^2}=\frac{1}{5}+\color{blue}{\text{infinite geometric progression}} $$

Let

$$\begin{align}f(x)&=\sum_{n=0}^\infty(2n+1)x^{2n+1}\\&=x\sum_{n=0}^\infty(2n+1)x^{2n}\\&=x\frac{d}{dx}\left(\sum_{n=0}^\infty x^{2n+1}\right)\\&=x\frac{d}{dx}\left( \frac{x}{1-x^2}\right)\\&=x\frac{x^2+1}{(1-x^2)^2}\end{align}$$ and notice that the desired sum is $f\left(\frac15\right)$.