(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 time \(t = n\) be \(T\). Then,