1 Question 1
Proof by induction:
When n = 1 we can not split the coin so the result is zero:
n(n 1)=2 = 1(1 1)=2 = 0 Base case hold (1)
Induction Step: Assume let k 2 N and k >= 1, the relation is true from the
base case n = 1 to n = k, which is: winning exactly k(k 1)=2 dollars.
We want to show that the relation holds when n = k + 1 using Induction
hypothesis. When n = k+1 we can split the coins in two piles, one with a coins
and one with b coins where a + b = k + 1.
Notice that a and b are both less or equal to k and greater than 1, therefore the
splitting of a and b should win k(k 1)=2 dollars base on our assumption. So
the total of winning from splitting n = k + 1 coins can be calculated as follow:
Since the relation still holds for n = k +1, by our IH, you always win a total of
exactly n(n 1)=2 dollars with n coins to start with.
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