In this question, you may use without proof the results
\[\sum_{r=0}^{n} \binom{n}{r} = 2^n \quad \text{and} \quad \sum_{r=0}^{n} r\binom{n}{r} = n\,2^{n-1}.\]
Show that
\[r\binom{2n}{r} = (2n+1-r)\binom{2n}{2n+1-r}\]
for \(1 \leqslant r \leqslant 2n\). Hence show that
\[\sum_{r=0}^{2n} r\binom{2n}{r} = 2\sum_{r=n+1}^{2n} r\binom{2n}{r}.\]
A fair coin is tossed \(2n\) times. The value of the random variable \(X\) is whichever is the larger of the number of heads and the number of tails shown. If \(n\) heads and \(n\) tails are shown, then \(X = n\).
Show that
\[\mathrm{E}(X) = n\left(1 + \frac{1}{2^{2n}}\binom{2n}{n}\right).\]
Show that \(\dfrac{1}{2^{2n}}\dbinom{2n}{n}\) decreases as \(n\) increases.
In a game, you choose a value of \(n\) and pay \(\pounds n\); then a fair coin is tossed \(2n\) times. You win an amount in pounds equal to the larger of the number of heads and the number of tails shown. If \(n\) heads and \(n\) tails are shown, then you win \(\pounds n\). How should you choose \(n\) to maximise your expected winnings per pound paid?