State the value of ${u_1}$.

Show that ${u_n}$ satisfies the recurrence relation

$${u_n} = \frac{1}{4}{u_{n - 1}} + \frac{1}{4}.$$

Solve this recurrence relation to find the probability that Andy wins the ${n^{{\text{th}}}}$ game.

After they have played many games, Pat comes to watch. Use your answer from part (c) to estimate the probability that Andy will win the game they are playing when Pat arrives.

$\frac{1}{2}$     A1

[1 mark]

Andy could win the ${n^{{\text{th}}}}$ game by winning the $n - {1^{{\text{th}}}}$ and then winning the ${n^{{\text{th}}}}$ game or by losing the $n - {1^{{\text{th}}}}$ and then winning the ${n^{{\text{th}}}}$     (M1)

${u_n} = \frac{1}{2}{u_{n - 1}} + \frac{1}{4}(1 - {u_{n - 1}})$     A1A1M1

Note:     Award A1 for each term and M1 for addition of two probabilities.

${u_n} = \frac{1}{4}{u_{n - 1}} + \frac{1}{4}$     AG

[4 marks]

general solution is ${u_n} = A{\left( {\frac{1}{4}} \right)^n} + p(n)$     (M1)

for a particular solution try $p(n) = b$     (M1)

$b = \frac{1}{4}b + \frac{1}{4}$     (A1)

$b = \frac{1}{3}$

hence ${u_n} = A{\left( {\frac{1}{4}} \right)^n} + \frac{1}{3}$     (A1)

using ${u_1} = \frac{1}{2}$     M1

$\frac{1}{2} = A\left( {\frac{1}{4}} \right) + \frac{1}{3} \Rightarrow A = \frac{2}{3}$

hence ${u_n} = \frac{2}{3}{\left( {\frac{1}{4}} \right)^n} + \frac{1}{3}$     A1

Note:     Accept other valid methods.

[6 marks]

for large $n{\text{ }}{u_n} \approx \frac{1}{3}$     (M1)A1

[2 marks]

Total [13 marks]

Not all candidates wrote this answer down correctly although it was essentially told you in the question.

Very badly answered. Candidates seemed to think that they were being told this relationship (so used it to find u(2)) rather than attempting to prove it.

This distinguished the better candidates. Some candidates though that they could use the method for homogeneous recurrence relations of second order and hence started solving a quadratic. Only the better candidates saw that it was a combined AP/GP.

The best candidates saw this but most had not done enough earlier to get to do this.