1. Finding the series

I have a series that goes by the form of:

1 + a + a^2 + a^3 +...+ a^n

where n is the number of terms in the series (not including the 1 at the beginning). So, for example, the series with 4 terms would take the form

1+ a + a^2 + a^3 + a^4

Is there a way to represent this in a non-geometric series type of way? More specifically, if I wanted a=3 and n=4, then the series looks like:

1 + 3 + 3^2 + 3^3 + 3^4 = 118

Is there a way to get 118 without a series representation and only knowing a=3 n=4?

2. Re: Finding the series

Originally Posted by colerelm1
I have a series that goes by the form of:

1 + a + a^2 + a^3 +...+ a^n

where n is the number of terms in the series (not including the 1 at the beginning). So, for example, the series with 4 terms would take the form

1+ a + a^2 + a^3 + a^4

Is there a way to represent this in a non-geometric series type of way? More specifically, if I wanted a=3 and n=4, then the series looks like:

1 + 3 + 3^2 + 3^3 + 3^4 = 118

Is there a way to get 118 without a series representation and only knowing a=3 n=4?
See here and look under the topic "formula."

-Dan

3. Re: Finding the series

Hello, colerelm1!

I have a series: . $S \:=\:1 + a + a^2 + a^3 + \hdots + a^n$

where $n$ is the number of terms in the series (not including the 1).

So, for example, the series with 4 terms would take the form: . $S_4 \:=\:1+ a + a^2 + a^3 + a^4$

Is there a way to represent this in a non-geometric series type of way?
What does that mean? .It is a geometric series. .How are we going to avoid that?

More specifically, if I wanted $a=3$ and $n=4$,
. . then the series looks like: . $S_4 \:=\:1 + 3 + 3^2 + 3^3 + 3^4 \:=\: \color{red}{121}$

Is there a way to get 121 without a series representation and only knowing a=3 n=4?

If you don't know the formula for the sum of a geometric series, you can derive it.

$\begin{array}{cccccc}\text{We have:} & S &=\; 1 + a + a^2 + a^3 + \hdots + a^n \qquad\quad \\ \text{Multiply by }a\!: & aS &=\; \qquad a + a^2 + a^3 + \hdots + a^n + a^{n+1} \end{array}$

$\text{Subtract: }\:S - aS \;=\;1 - a^{n+1} \quad\Rightarrow\quad (1-a)S \:=\:1-a^{n+1}$

Therefore: . $S \:=\:\frac{1-a^{n+1}}{1-a} \quad\Rightarrow\quad \boxed{S \;=\;\frac{a^{n+1}-1}{a-1}}$