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Math Help - geometric series help.

  1. #1
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    geometric series help.

    hello

    i am currently studying geometric series. and i have been trying to do some questions which i just can't seem to solve. i have scanned the questions and have attached them to this thread.

    can someone please help

    i need to q5 and q6

    i did attempt the questions, however i was getting incorrect answers.

    any help is appreichated

    thanks

    regards
    Attached Thumbnails Attached Thumbnails geometric series help.-mathq-s.jpg  
    Last edited by rpatel; January 8th 2007 at 08:56 AM.
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  2. #2
    TD!
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    Try to get them in a standard form, for which you probably saw a formula:

    <br />
\sum\limits_{r = 1}^\infty  {\frac{{3k}}{{6^r }}}  = 3k\sum\limits_{r = 1}^\infty  {\left( {\frac{1}{6}} \right)^r }  = 3k\left( {\sum\limits_{r = 0}^\infty  {\left( {\frac{1}{6}} \right)^r  - 1} } \right) = 3k\left( {\frac{1}{{1 - \frac{1}{6}}} - 1} \right) = \frac{3}{5}k<br />
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  3. #3
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    when u mean formula do mean

    a(r^n-1)
    ----------
    r-1

    and


    a(1-r^n)
    ----------
    1-r


    ?

    thanks

    regards
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  4. #4
    TD!
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    For an infinite sum, we even have something simpler. I used:

    \sum\limits_{r = 1}^\infty  {a^r }  = \frac{1}{{1 - a}}
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  5. #5
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    oh ok.

    can you please just help me out on any question from q6 please

    thanks

    regards
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  6. #6
    TD!
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    Notice that:

    <br />
\frac{9}{4} = \left( {\frac{2}{3}} \right)^{ - 2} ,\frac{3}{2} = \left( {\frac{2}{3}} \right)^{ - 1} ,0 = \left( {\frac{2}{3}} \right)^0 , \cdots ,\frac{{1024}}{{59019}} = \left( {\frac{2}{3}} \right)^{10} <br />

    So if we shift the exponent by 2, we start at exponent 0 and go to 12.
    We can then write this finite geometrical sum compactly, as follows:

    <br />
S = \sum\limits_{n =  - 2}^{10} {\left( {\frac{2}{3}} \right)^n }  = \frac{{\left( {\frac{2}{3}} \right)^{ - 2}  - \left( {\frac{2}{3}} \right)^{11} }}{{1 - \frac{2}{3}}} = \frac{{1586131}}{{236196}} \approx 6.72<br />
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  7. #7
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    when i was doing this i was using logs.

    but i ended up with wrong answer

    TD. do you think i should do exponentials and logarithms before attempting geometric sequences ?
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  8. #8
    TD!
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    Do you understand what I did? And do you recognise the formula I used?
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  9. #9
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    the formula you use was

    a(1-r^n)
    ----------
    1-r


    am i right in saying this ?

    thanks

    regards
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  10. #10
    TD!
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    Almost, that's for when your index starts at 0, now we began at -2.
    The first 1 in the numerator is now r^q instead of 1, where q is this start index.
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  11. #11
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    is there a way of doing q6 with logarithms

    the reason why i ask is because the examples i came accross used logs.
    and the formula used the answer above is one that i am unaware of and also its not in my textbook

    however the answer is correct though.

    many thanks

    regards
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  12. #12
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    Quote Originally Posted by rpatel View Post
    hello

    i am currently studying geometric series. and i have been trying to do some questions which i just can't seem to solve. i have scanned the questions and have attached them to this thread.

    can someone please help

    i need to q5 and q6

    i did attempt the questions, however i was getting incorrect answers.

    any help is appreichated

    thanks

    regards
    Here is one way.

    [Q5,a] Summation of [3k / 6^n] from n=1 to n=infinity.
    I changed r into n because I want to maintain r = ratio
    In series form, that is
    3k/(6^10 +3k/(6^2) +3k/(6^3) +....+3k/(6^infinity)

    r = a2/a1 = 6^(-1) = 1/6

    It is a geometric series where r = 1/6
    So,
    Sn = (a1)[(1-r^n)/(1-r)]
    n = infinity
    S(infinity) = (3k /6)[(1 -(1/6)^infinity) / (1 -1/6)]
    = (3k /6)[(1-0) / (5/6)]
    = (3k /6)[6/5]
    = 3k /5 ------------------answer.

    [Q5,b] Summation of [(-2)^n +n -2] from n=3 to n=7.
    Split that into a geometric and an arithmetic series.
    Sn = (a1)[(1-r^n)/(1-r)] +(n/2)[b1 +bn]

    In the geometric series (-2)^n, the r is (-2).
    In the arithmetic series (n-2), the d is 1.

    n=3 to n=7 is 5 elements.

    So,
    S(5) = (-2)^3 *[(1 -(-2)^5)/(1-(-2))] +(5/2)[(3-2) +(7-2)]
    S(5) = (-8)[(1-(-32)/(1+2)] +(2.5)[6]
    S(5) = -88 +15
    S(5) = -73 ---------answer

    Note : S(infinity) and S(5) read "S sub infinity" and "S sub 5".

    -----------------------------------------------
    [Q6,a]

    9/4 +3/2 +1 +.....+1024/59,049

    A geometric series.
    r = (3/2)/(9/4) = 2/3

    a1 = 9/4 = (3/2)^2 = (2/3)^(-2)
    an = 1024/59,049 = (2/3)^10
    So from -2 to 10. That's 13 elements.

    S(13) = (9/4)[(1 -(2/3)^13) / (1 -(2/3)]
    = (9/4)[(0.99486)/(0.33333)]
    = 6.71537
    = 6.72 --------------------answer.

    Sorry, no more time, gotta go.
    Last edited by ticbol; January 9th 2007 at 12:40 AM. Reason: 3 to 7 is 5, not 4. Then, (1-(-2)) is 3, not 4. Phew!
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  13. #13
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    Quote Originally Posted by rpatel View Post
    hello

    i am currently studying geometric series. and i have been trying to do some questions which i just can't seem to solve. i have scanned the questions and have attached them to this thread.

    can someone please help

    i need to q5 and q6

    i did attempt the questions, however i was getting incorrect answers.

    any help is appreichated

    thanks

    regards
    Let me finish this question.

    [Q6,b]
    1/5 -3/20 +9/80 -....+6561/327,680

    r = (-3/20) / (1/5) = (-3/20)*(5/1) = -3/4
    r = (9/80) /(-3/20) = (9/80)*(-20/3) = -3/4

    Okay, so the series is geometric, with r = -3/4 and a1 = 1/5

    Rewriting the series,
    (1/5)(-3/4)^0 +(1/5)(-3/4)^1 +(1/5)(-3/4)^2 +....+(1/5)(-3/4)^8
    From 0 to 8, so there are 9 elements, or n=9.

    S(9) = (1/5)[(1 -(-3/4)^9) / (1 -(-3/4))]
    = (1/5)[(1.075)/(1.75)]
    = 0.123 ----------------------answer.
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  14. #14
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    thank you

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