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

I am a high school math teacher who likes to understand the development and rationale behind formulas, albeit I am not a math expert by any means.

Could I get some help in trying to explain how the formaula for the sum of a finite geometric series actually can be derived? I need a a rather simple explanation initially, so that I can grasp the essential nature of the formula. Thanks for allowing me some space here.

• Like $$(1-q)\sum_{k=0}^n q^k = \sum_{k=0}^n q^k - \sum_{k=1}^{n+1}q^k = 1 - q^{n+1}\,?$$ – Daniel Fischer May 8 '14 at 17:50
• this has been answered a million billion times before elsewhere on the site – enthdegree May 8 '14 at 17:54
• For a handout I used to use that may be appropriate, see the file titled "seq-and-series.pdf" that I posted here. For a more advanced treatment with applications, see the file titled "geom-growth.pdf" at the same place. – Dave L. Renfro May 8 '14 at 18:08

How about $$s=a+ar+ar^2+\dots ar^n\\rs=ar+ar^2+ar^3+\dots ar^{n+1}\\(r-1)s=ar^{n+1}-a\\s=\frac{ar^{n+1}-a}{r-1}=a\frac{r^{n+1}-1}{r-1}$$
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 $$a_0\frac{r^{n+1}-1}{r-1}$$