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Math Help - series with geometric terms, find the sums.

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

    There are two things to note.

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

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

    In our case, the first term is a=5. Thus, the series \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 \sum_{n=0}^{\infty}5\left(-\tfrac{1}{4}\right)^n\sim\sum_{n=0}^{\infty}ar^n, we see that {\color{red}a=5} and {\color{blue}r=-\tfrac{1}{4}}

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

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

    This time, though, we see that it expands to \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 ar, thus it converges to \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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