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Thread: series with geometric terms, find the sums.

  1. #1
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    series with geometric terms, find the sums.

    here is what i did. i guessu gotta know the formula (a)/(1-r) to find the sum. here is what i did no idea if i m right. Wondering if i could get some help on how this is done thanks. to my understand a is the first value of the series and r is related to the n.. not sure.

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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    here is what i did. i guessu gotta know the formula (a)/(1-r) to find the sum. here is what i did no idea if i m right. Wondering if i could get some help on how this is done thanks. to my understand a is the first value of the series and r is related to the n.. not sure.

    The series $\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{5}{4^n}$ is the same as $\displaystyle \sum_{n=0}^{\infty}5\left(-\frac{1}{4}\right)^n$, which has the form of a geometric series $\displaystyle \sum_{n=0}^{\infty}ar^n$

    There are two things to note.

    When you expand the series and it starts off like $\displaystyle a+ar+ar^2+\dots$, then the series sums up to $\displaystyle \boxed{\frac{a}{1-r}}$.

    However, if you expand the series, and it starts off like $\displaystyle ar+ar^2+ar^3+\dots$, then the series sums up to $\displaystyle \frac{a}{1-r}-a=\boxed{\frac{ar}{1-r}}$.

    In our case, the first term is $\displaystyle a=5$. Thus, the series $\displaystyle \sum_{n=0}^{\infty}5\left(-\frac{1}{4}\right)^n=\frac{5}{1+\frac{1}{4}}=\colo r{red}\boxed{4}$

    You're answer was correct. Does this make sense?
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  3. #3
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    kind of but how do u know which way it expands

    this series would expand like this right?:

    (5) + (1/4) + (5/16) .......

    how do i know if that is ar + ar^2 or a + ar ??
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    kind of but how do u know which way it expands

    this series would expand like this right?:

    (5) + (1/4) + (5/16) ....... Almost....take into account that there will be alternating signs.

    how do i know if that is ar + ar^2 or a + ar ??
    First, you identify a and r from the geometric series.

    Since our series was $\displaystyle \sum_{n=0}^{\infty}5\left(-\tfrac{1}{4}\right)^n\sim\sum_{n=0}^{\infty}ar^n$, we see that $\displaystyle {\color{red}a=5}$ and $\displaystyle {\color{blue}r=-\tfrac{1}{4}}$

    In expanding this series, we have $\displaystyle \underbrace{{\color{red}5}}_{\color{red}a}-\tfrac{5}{4}+\tfrac{5}{16}-\tfrac{5}{64}+\dots$. The first term in our series is $\displaystyle a$, thus we say it converges to $\displaystyle \frac{a}{1-r}\implies\frac{5}{1+\frac{1}{4}}=\color{red}\boxe d{4}$

    However, what if our series was $\displaystyle \sum_{n=1}^{\infty}5\left(-\tfrac{1}{4}\right)^n$? We see that its similar to $\displaystyle \sum_{n=1}^{\infty}ar^n$, and can conclude [again] that $\displaystyle {\color{red}a=5}$ and $\displaystyle {\color{blue}r=-\tfrac{1}{4}}$.

    This time, though, we see that it expands to $\displaystyle \underbrace{{\color{blue}-}\tfrac{{\color{red}5}}{{\color{blue}4}}}_{{\color {red}a}{\color{blue}r}}+\tfrac{5}{16}-\tfrac{5}{64}+\dots$. The first term in our series is $\displaystyle ar$, thus it converges to $\displaystyle \frac{a}{1-r}-a=\frac{ar}{1-r}\implies\frac{5\left(-\tfrac{1}{4}\right)}{1+\frac{1}{4}}=\color{red}\bo xed{-1}$

    Does this clarify things??
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