<p>A checkers tournament has 128 entries. When a player loses a game, he or she is eliminated. Winners play in the next round. How many games are played before a winner is determined?</p>
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<p>I think I’m crazy or just really tired, because I got 65 and the answer is 127. This is geometric series problem in Pre-calc 11 by the way so I would appreciate if you could show me how you solved it using a series formula if you can :D</p>
<p>thanks!</p>
<p>There’s actually a solution to this that uses logic. Look:</p>
<p>For every game that is played, one player loses.
Out of 128 players, 1 won
Therefore 127 players lost
Therefore, 127 games were played.</p>
<p>As for series, meh, someone else do it.</p>
<p>It is 127 because you have 64 games going on (two people paired up) then the next roun you have 32 games then the next round 16 then the next round 8 then the next round 4 then the next round 2 and then the final game there is one game and add that all together you get 127 games.
Eh…using series idk I just used logic</p>
<p>Nevermind! Stupid question! </p>
<p>But I would still appreciate if someone could show me the way to do it using a series formula…</p>
<p>I’m not sure for series it’d probably be like (n/2) plus eh I have nooo clue :(</p>
<p>Yeah, you have a geometric series.</p>
<p>S = a(1-r^n)/(1-r)
Where n = lb(128) = 7, this is not an infinitely convergent series, because it stops when there is only 1 game. You have 2 players per game, and 128 players total, for 7 “levels”.
a = first term in the set, which is 64, because 128 people play 64 games.
r = 1/2 because every time, one player of every 2 loses</p>
<p>S = 64(1-1/2^7)/(1-1/2) = 127</p>
<p>You can also use the sigma function for 2^n. Use n=6, as 2^7 is 128, therefore you should use 6.</p>
<p>1/2^7 is not 128, it’s 1/128
You don’t have to use the sigma notation when you can use a binary logarithm.</p>