Sum of a Finite Geometric series is hard for me to explain to my high school students. Is there a simple explanation?

How about $$s=a+ar+ar^2+\dots ar^n\\rs=ar+ar^2+ar^3+\dots ar^{n+1}\\(r-1)s=ar^{n+1}-a\\s=\frac{ar^{n+1}-a}{r-1}=a\frac{r^{n+1}-1}{r-1}$$

Try with a proof wihout words:

http://www.cut-the-knot.org/Curriculum/Algebra/SequencesSquare.shtml

I don't know if you or your students are familiar with bases, but I like to think about it this way:

In base $10$, note that $$\begin{align}9+90&=99=100-1 \\ 9+90+900&=999=1000-1 \\ 9+\cdots +9000&=10000-1\end{align}$$ So in general it seems that $$9+\cdots +9\times10^n=10^{n+1}-1$$ Or in base $2$, we get $$\begin{align}1+10=&100-1 \\ 1+10+100=&1000-1 \\ \vdots\end{align}$$ So in general, for base $r$ it holds that $$(r-1)+r(r-1)+r^2(r-1)+\cdots r^n(r-1)=r^{n+1}-1$$ Therefore $$(r-1)(1+r+\cdots r^n)=r^{n+1}-1\implies 1+r+\cdots r^n=\frac{r^{n+1}-1}{r-1}$$ If there is a constant in front, just factor it out to get $$a_0\frac{r^{n+1}-1}{r-1}$$