Question 1153615
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If there are n teams, G(n) = n(n-1)/2

If there are m=2n teams, G(m) = m(m-1)/2 = 2n(2n-1)/2 = n(2n-1) games

The difference is
{{{ n(2n-1) - n(n-1)/2 = (2n(2n-1) - n(n-1))/2 = (3n^2-n)/2 }}}

So, NO, the number of games does not double, it increasses by {{{ (n(3n-1))/2 }}} when n doubles.
 

Example:

G(6) = 6(5)/2 = 15
G(12) = 12(11)/2 = 66

The difference is 66-15 = 51  
and  for n=6: n(3n-1)/2 = 6(3*6-1)/2 = 6(17)/2 = 51, as derived above.

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To the student:  sorry just noticed your additional question in the 'thank you' message:  it means when you double the number of teams, the number of games goes up by {{{ (3n^2-n) /2 }}} or we can say (roughly) "if the number of teams doubles, the number of games needed goes up on the order of {{{n^2}}}".  In the example I gave, n=6, G(n)=15  then I doubled the number of teams to 12 and showed G(12) = 66 so you can see the number games went up by a factor of 66/15 or 4.4.  4.4 is _roughly_ (on the same order of magnitude as) {{{2^2}}}

Forgot to say... so yes, for doubling teams, you roughly quadruple the # of games needed for the RR.