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Math Help - Geometric Sequences and Series

  1. #1
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    Geometric Sequences and Series

    Q. Find Sn, the sum to n terms, of the geometric series 2 + 2/ 3 + 2/ 3^2 +…
    If Sn = 242/ 81, find the value of n.

    Eq. Sn = a (1 - r^n) / (1 - r) … (r < 1).

    Attempt:
    From the series above, r = (2/ 3) / 2 => 1/ 3
    Therefore, a = 2 and r = 1/ 3. Sub these into the Sn equation…
    2 (1 - (1/ 3)^n) / (1 - 1/3) =
    2 (1 - (1/ 3)^n) / (2 / 3)

    Answer:
    I’m stuck at this point, as the answer in the text book indicates that:

    Sn = 3 - (1/ (3^(n-1)))

    I am uncertain of the procedure they have applied to convert n into n - 1. Can anyone help? Thank you.
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  2. #2
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    Re: Geometric Sequences and Series

    Your working looks good, to finish it,

    Solve for n where \displaystyle \frac{242}{81} = \frac{1-\frac{1}{3}^n}{\frac{2}{3}}
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  3. #3
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    Re: Geometric Sequences and Series

    Quote Originally Posted by GrigOrig99 View Post
    Q. Find Sn, the sum to n terms, of the geometric series 2 + 2/ 3 + 2/ 3^2 +…
    If Sn = 242/ 81, find the value of n.

    Eq. Sn = a (1 - r^n) / (1 - r) … (r < 1).
    ...

    Answer:
    I’m stuck at this point, as the answer in the text book indicates that:

    Sn = 3 - (1/ (3^(n-1)))

    I am uncertain of the procedure they have applied to convert n into n - 1. Can anyone help? Thank you.
    The 1st term in the series is (2)(1/3)^0 . It has an exponent of 0.

    The 2nd is (2)(1/3)^1. It has an exponent of (2-1).
    \vdots
    The n-th term is 2)(1/3)^(n-1). It has an exponent of (n-1).
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    Re: Geometric Sequences and Series

    Ok, so following on from:
    2 (1 - (1/ 3)^n) / (2 / 3)

    I have done the following:

    = (2(1) - 2(1/ 3)^n) / (2/ 3)
    = (2 - (2/3)^n) / (2/ 3)
    = (2(3/ 2) - (3/ 2)((2/ 3)^n))
    = (6/ 2 - (6/ 6)^n)
    = 3 - 1^(n-1)

    Recall that the text book answer is Sn = 3 - (1/ (3^(n-1))).

    I guess I'm closer to the mark now, but I still don't have it entirely correct. Can anyone see where I've gone wrong? Thanks again.
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  5. #5
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    Re: Geometric Sequences and Series

    Quote Originally Posted by GrigOrig99 View Post
    Ok, so following on from:
    2 (1 - (1/ 3)^n) / (2 / 3)

    I have done the following:

    = (2(1) - 2(1/ 3)^n) / (2/ 3)
    You're good to the above line.

    = (2 - (2/3)^n) / (2/ 3)
    (2)(1/3)^n ≠ (2/3)^n)

    = (2(3/ 2) - (3/ 2)((2/ 3)^n))
    = (6/ 2 - (6/ 6)^n)
    = 3 - 1^(n-1)

    Recall that the text book answer is Sn = 3 - (1/ (3^(n-1))).

    I guess I'm closer to the mark now, but I still don't have it entirely correct. Can anyone see where I've gone wrong? Thanks again.
    2\ \left(\frac{1 - (1/ 3)^n}{2 / 3}\right)= \frac{2\cdot 3(1 - (1/ 3)^n)}{2}\right=3-(1/3)^{(n-1)}

    Notice that \frac{242}{81}=\frac{243}{81}-\frac{1}{81}=3-\frac{1}{81}
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  6. #6
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    Re: Geometric Sequences and Series

    Ok, I'm almost clear on this. So, is it correct to say that for:

    (2.3 (1 - (1/3)^n)/ 2

    we do 'not' actually multiply a number (1/3^n, in this case) by another (i.e., 2.3 or 6) when it is expressed to the power of an unknown (n)?
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  7. #7
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    Re: Geometric Sequences and Series

    Yes, but we can mulitply them if the power's base is the same as the constant. For example: 3\cdot 3^x=3^{x+1}. (I don't know if this answers your question).

    Have you found the value of n now? ...
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  8. #8
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    Re: Geometric Sequences and Series

    Great, thanks for clearing that up. n should be 5, then.

    Thanks again to everyone who helped.
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  9. #9
    MHF Contributor Siron's Avatar
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    Re: Geometric Sequences and Series

    Right! n=5
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