# Geometric sums and series.

• Mar 19th 2009, 11:32 AM
brentwoodbc
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.
• Mar 19th 2009, 01:29 PM
GP
Hello brentwoodbc
Quote:

Originally Posted by brentwoodbc
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:

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

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

• Mar 19th 2009, 01:55 PM
brentwoodbc
thanks
Quote:

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:

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

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

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
$
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.
• Mar 19th 2009, 10:39 PM
Quote:

Originally Posted by brentwoodbc
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
$
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!