Can someone please help with part (iii) for this question? Thank you.

Edit: At later glance I realised I slightly messed up the base.\[ \text{When }n=0,\\ \text{the only configurations for }(r,s)\text{ are }(1,0)\text{ and }(0,1).\\ \text{So we can just check each individually.}\]

\[ \text{For the }(0,1)\text{ case, we have }P(0,1)=0.\\ \text{Here }s=1\text{, so we verify that}\\ \frac{1}{2^1}\left[ \binom10 + \binom11 \right] =\frac22 = 1 \]

\[ \text{For the }(1,0)\text{ case, we have }P(1,0) = 0.\\ \text{Here }s=0\text{, which is outside the valid range }s\geq 1.\\ \text{Hence this does not contradict the inductive assumption anyhow.} \]

So in both cases, the base case \(n=0\) is valid.

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\[ \text{Now assume the statement holds }\textbf{for all}\text{ configurations }(r,s)\\ \text{such that }r+s-1=k. \\ \text{We then need to prove the statement holds for all configurations }(r,s)\\ \text{such that } r+s-1=k+1. \]

That was perhaps the most confusing part of the induction - figuring out the goal of the inductive step.

\[ \text{Firstly, from the definition of the recurrence,}\\ P(r,s) = \frac12 P(r-1, s) + \frac12 P(r,s-1). \]

\[ \text{Now if }r+s-1 = k+1\text{, we know that}\\ (r-1)+s-1 = k\text{ and } r + (s-1) - 1= k.\\ \text{Hence we may use the inductive assumption on }\textit{both}\text{ terms:} \]

\begin{align*}

P(r,s) &= \frac12 P(r-1, s) + \frac12 P(r,s-1)\\

&= \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] + \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right] \\

&= \frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] +\frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right]

\end{align*}

Ensure that you understand why the first sum terminates at \(s-1\), whilst the second terminates at \(s-2\).

\[ \text{Now to finish, the tricks here are similar to the induction proof of the binomial theorem.}\\ \text{We need to recall that firstly, }\binom{k}{0} = \binom{k+1}{0} = 1.\\ \text{Also, we need Pascal's identity }\boxed{\binom{N+1}{K+1} = \binom{N}{K} + \binom{N}{K+1}}.\\ \text{With }\textit{both}\text{ properties in mind, we proceed to cleverly rearrange the terms.} \]

\begin{align*}

P(r,s) &= \frac1{2^{n+1}} \left[ \binom{k}{0} + \left[ \binom{k}0 + \binom{k}1\right] + \left[ \binom{k}1+ \binom{k}2 \right]+ \dots + \left[\binom{k}{s-2}+\binom{k}{s-1} \right] \right]\\

&= \frac1{2^{n+1}} \left[ \binom{k+1}{0} + \binom{k+1}{1}+\binom{k+1}{2} + \dots + \binom{k+1}{s-1} \right].

\end{align*}

And this is exactly what we wished to prove, so we are done.

*I did this in a bit of a rush so I may have given too little clarification. Let me know if there's any bit you want me to elaborate on