Answer:
Step-by-step explanation:
given that in a round robin tennis tournament, each player plays every other player exactly once.
Suppose there are two players i.e. n=2, we have only one match satisfies
[tex]\frac{2(2-1)}{2} =1[/tex]
Hence P(2) is true
Assume P(k) is true. For k players no of matches played
= [tex]\frac{k(k-1)}{2}[/tex]
To prove true for n = k+1
If to k players one new player is introduced, then the new player should play all the k players to have the condition satisfied
i.e. no of matches = no for k players + k
= [tex]\frac{k(k-1)}{2} +k\\= \frac{k^2-k+2k}{2} \\= \frac{(k+1)k}{2}[/tex]
So if true for n =k, then true for n =k+1
Already true for n =2
By induction true for all natural numbers starting from 2.
