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Math Help - [SOLVED] Finding The Sum Of This Series

  1. #1
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    [SOLVED] Finding The Sum Of This Series

    I've attached the problem I'm stuck on, I tried using the formula for geometric series several times but I'm guessing that because there are two terms I should be going about it a different way? Could someone help me out?
    Attached Thumbnails Attached Thumbnails [SOLVED] Finding The Sum Of This Series-picture-1.png  
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  2. #2
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    Quote Originally Posted by fattydq View Post
    I've attached the problem I'm stuck on, I tried using the formula for geometric series several times but I'm guessing that because there are two terms I should be going about it a different way? Could someone help me out?
    Based on the answer you gave, it looks like you were close. The key thing to notice is that the exponent on \frac{1}{2} is n-1. Thus, you sum is equivalent to \sum_{n=0}^{\infty}7\left(\frac{1}{2}\right)^n

    The rest of the solution:
    Spoiler:

     = 7\cdot\frac{1}{1-\frac{1}{2}} = 14
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    Based on the answer you gave, it looks like you were close. The key thing to notice is that the exponent on \frac{1}{2} is n-1. Thus, you sum is equivalent to \sum_{n=0}^{\infty}7\left(\frac{1}{2}\right)^n

    The rest of the solution:
    Spoiler:

     = 7\cdot\frac{1}{1-\frac{1}{2}} = 14
    How can you just change n to 0? You can do that?
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  4. #4
    Super Member redsoxfan325's Avatar
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    I also changed the exponent from n-1 to n. In the previous sum, n starts at 1 so n-1 starts at 0. In this sum, both n and the exponent start at 0. It's the same. If you have doubts, try writing out the first few terms, and you'll see that they're the same.
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  5. #5
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    Hello, fattydq!

    Find the sum of the series if it converges: . \sum^{\infty}_{n=1}7\left(\tfrac{1}{2}\right)^{n-1}

    ... but I'm guessing that because there are two terms ... . .
    What two terms?

    We have: . S \;=\;7 + 7\left(\tfrac{1}{2}\right) + 7\left(\tfrac{1}{2}\right)^2 + 7\left(\tfrac{1}{2}\right)^3 + \hdots

    This is an infinite geometric series with first term a = 7 and common ratio r = \tfrac{1}{2}

    Its sum is: . S \;=\;\frac{7}{1 - \frac{1}{2}} \;=\;14


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    That second problem is interesting . . .


    \sum^{\infty}_{n=0}\frac{(6\pi)^n}{18^{n+1}}

    We have: . \frac{(6\pi)^n}{18^{n+1}} \;=\;\frac{(6\pi)^n}{18\cdot18^n} \;=\; \frac{1}{18}\cdot\frac{(6\pi)^n}{18^n} \;=\;\frac{1}{18}\left(\frac{6\pi}{18}\right)^n \;=\;\frac{1}{18}\left(\frac{\pi}{3}\right)^n


    And the series is: . S \;=\;\frac{1}{18} + \frac{1}{18}\left(\frac{\pi}{3}\right) + \frac{1}{18}\left(\frac{\pi}{3}\right)^2 + \frac{1}{18}\left(\frac{\pi}{3}\right)^3 +\hdots


    This too is an infinite geometric series,

    . . but since  r \,=\,\tfrac{\pi}{3}\,>\,1, the series diverges.

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