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Math Help - Geometric sums and series.

  1. #1
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    Question Geometric sums and series.

    64 players enter a tennis tournament. When a player loses a match the player drops out, the winners go onto the next round.

    Find as many different methods as you can to determine the number of matches played.



    I did it using Sum6=(32(.5^{6}-1))/(.5-1)

    which equal 63 which is right... But I got n=6 by finding that there are 6 rounds by going - 64/2=32 32/2=16 etc.

    What is the correct way to solve this say if there was 10, 000 rounds and you didnt want to waste time. thank you.
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  2. #2
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    GP

    Hello brentwoodbc
    Quote Originally Posted by brentwoodbc View Post
    64 players enter a tennis tournament. When a player loses a match the player drops out, the winners go onto the next round.

    Find as many different methods as you can to determine the number of matches played.



    I did it using Sum6=(32(.5^{6}-1))/(.5-1)

    which equal 63 which is right... But I got n=6 by finding that there are 6 rounds by going - 64/2=32 32/2=16 etc.

    What is the correct way to solve this say if there was 10, 000 rounds and you didnt want to waste time. thank you.
    To find the number of terms ( n) of a GP, if you know the first and last terms, use the fact that the n^{th} term is ar^{n-1}, where a is the first term and r is the common ratio.

    For example, we can find the number of terms in the GP 3 + 15 + 75 + ... + 732421875, by saying a = 3, r = 5 and ar^{n-1} = 732421875

    \Rightarrow 3 \times 5^{n-1} = 732421875

    \Rightarrow 5^{n-1} = \frac{732421875}{3} = 244140625

    Now take logs (to any base) of both sides:

    (n-1)\log 5 = \log 244140625

    \Rightarrow n = 1 + \frac{\log 244140625}{\log 5} = 13

    Does that answer your question?

    Grandad
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  3. #3
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    thanks
    Quote Originally Posted by Grandad View Post
    Hello brentwoodbcTo find the number of terms ( n) of a GP, if you know the first and last terms, use the fact that the n^{th} term is ar^{n-1}, where a is the first term and r is the common ratio.

    For example, we can find the number of terms in the GP 3 + 15 + 75 + ... + 732421875, by saying a = 3, r = 5 and ar^{n-1} = 732421875

    \Rightarrow 3 \times 5^{n-1} = 732421875

    \Rightarrow 5^{n-1} = \frac{732421875}{3} = 244140625

    Now take logs (to any base) of both sides:

    (n-1)\log 5 = \log 244140625

    \Rightarrow n = 1 + \frac{\log 244140625}{\log 5} = 13

    Does that answer your question?

    Grandad
    for the terms would you use 32 ie 32 original matches, or the 64 for the number of players? I guess it wouldn't matter as long as you knew what you used and corrected for the answer?


    so if you use 32 for a
    and r= 1/2
    so the final term is what the equation is equal to?

    like
    <br />
tn=ar^{n-1}

    1=32 x .5^{n-1}


    doing that I get n=6 which is the number of rounds.

    then I use the sum formula

    s6= (a(r^n-1)/(r-1)

    =63 right. Ok thank you.
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  4. #4
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    Quote Originally Posted by brentwoodbc View Post
    thanks


    for the terms would you use 32 ie 32 original matches, or the 64 for the number of players? I guess it wouldn't matter as long as you knew what you used and corrected for the answer?


    so if you use 32 for a
    and r= 1/2
    so the final term is what the equation is equal to?

    like
    <br />
tn=ar^{n-1}

    1=32 x .5^{n-1}


    doing that I get n=6 which is the number of rounds.

    then I use the sum formula

    s6= (a(r^n-1)/(r-1)

    =63 right. Ok thank you.
    Correct!

    Grandad
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