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Math Help - Finding the series

  1. #1
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    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?
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  2. #2
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    Re: Finding the series

    Quote Originally Posted by colerelm1 View Post
    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
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  3. #3
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    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}}
    Thanks from colerelm1
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