0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Like $$(1-q)\sum_{k=0}^n q^k = \sum_{k=0}^n q^k - \sum_{k=1}^{n+1}q^k = 1 - q^{n+1}\,?$$ $\endgroup$ – Daniel Fischer May 8 '14 at 17:50
  • 2
    $\begingroup$ this has been answered a million billion times before elsewhere on the site $\endgroup$ – enthdegree May 8 '14 at 17:54
  • 1
    $\begingroup$ 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. $\endgroup$ – Dave L. Renfro May 8 '14 at 18:08
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

Try with a proof wihout words:

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

$\endgroup$
1
$\begingroup$

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 $$a_0\frac{r^{n+1}-1}{r-1}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.