3
$\begingroup$

The formula to sum the terms of a finite geometric series is the following:

$$\frac{a_1(1 - r^{n+1})}{1 - r}$$

where $a_1$ is the first term, $r$ is the common ratio, and $n + 1$ is the number we want to sum up to.

Now, my problem is really in this last part, I have seen some formulas that use $r^n$, others use $r^{n+1}$.

My questions are:

  • In general, how does the power to which we raise $r$ change depending on the index of the sum?

  • What if we want to start the sum from a different index, for example $1$ or $3$ instead of $0$.

  • How does the number of terms we want to sum influence the formula?

I know these might seem like stupid questions, but I am just confused, and it might be the time to understand exactly what's going on.

I have seen the derivation of the formula, but I am still not understanding the indices.

$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

Observe

$$S_n=\sum_{i=m}^{n}r^n=r^m+r^{m+1}+\cdots +r^{n}\tag 1$$

$$rS_n=\sum_{i=m}^{n}r^n=r^{m+1}+\cdots r^{n}+r^{n+1}\tag 2$$

So, we see that $(1)$ and $(2)$ differ by only the first term of $(1)$ and the last term of $(2)$.

Thus, we have $(1-r)S_n=r^{m}-r^{n+1}$ whereupon dividing by $1-r$ reveals

$$S_n=\frac{r^m-r^{n+1}}{1-r}$$

$\endgroup$
2
$\begingroup$

In general, for $n\geq m$ and $r\neq 1$, $$ \sum_{k=m}^n r^k = r^m \sum_{k=m}^n r^{k-m} = r^m \sum_{\ell=0}^{n-m} r^{\ell} = r^m\cdot \frac{1-r^{n-m+1}}{1-r} $$ To remember it: the exponent in the denominator is the number of terms in the sum.

To check (very basic check): when $n=m$, you only have one term, and it's $r^m$—so the result should simplify to $r^m$ in that case.

$\endgroup$

You must log in to answer this question.