# Finding the series

• Nov 29th 2012, 01:20 PM
colerelm1
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?
• Nov 29th 2012, 01:48 PM
topsquark
Re: Finding the series
Quote:

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
• Nov 29th 2012, 02:25 PM
Soroban
Re: Finding the series
Hello, colerelm1!

Quote:

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

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

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

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

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.

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

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

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