# P.o.t.W. #14 Solution

(a) Let the total number of coins in the well at time $$t = n$$ be S. Then,$$1 + 2 + 3 + \ldots + \left( {n - 2} \right) + \left( {n - 1} \right) + n = S$$ (1)This can be re-written, such that$$n + \left( {n - 1} \right) + \left( {n - 2} \right) + \ldots + 3 + 2 + 1 = S$$ (2)$$\left( 1 \right) + \left( 2 \right)$$ gives:$$1 + n + \left[ {2 + \left( {n - 1} \right)} \right] + \left[ {3 + \left( {n - 2} \right)} \right] + \ldots + \left[ {\left( {n - 2} \right) + 3} \right] + \left[ {\left( {n - 1} \right) + 2} \right] + n + 1 = 2S$$$$\left( {n + 1} \right) + \left( {n + 1} \right) + \left( {n + 1} \right) + \ldots = 2S$$ (3)There are $$n$$ terms on LHS of (3), so$$n\left( {n + 1} \right) = 2S$$Thus,$$S = \frac{1}{2}n\left( {n + 1} \right)$$ Q.E.D.(b) Firstly, let us consider the scenario where no coins are removed from the well. In this case, let the total number of coins in the well at...

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