Use the formula for the sum of a finite geometric series to show that

\[{\text{P}}({X_n} = k) = \left\{ {\begin{array}{*{20}{l}} {\frac{1}{n}}&{{\text{for }}1 \leqslant k \leqslant n} \\ 0&{{\text{otherwise}}} \end{array}.} \right.\]

Find \({\text{E}}({X_n})\).

Find the set of values of \(n\) for which \({\text{E}}({X_{n - 1}} \times {X_{n + 1}}) < 2n\).

using \(\left( {\frac{{{t^n} - 1}}{{t - 1}}} \right) = 1 + t + {t^2} +  \ldots {t^{n - 1}}\)     M1

\({G_n}(t) = 0 + \frac{t}{n} + \frac{{{t^2}}}{n} + \frac{{{t^3}}}{n} +  \ldots \frac{{{t^n}}}{n} + 0 \times {t^{n + 1}} + 0 \times  \ldots \)    A1A1

Note:     A1 for the non-zero terms, A1 for the observation that all other terms are zero.

the statement that the coefficient of \({t^k}\) gives \({\text{P}}({X_n} = k)\)     R1

hence \({\text{P}}({X_n} = k) = \left\{ {\begin{array}{*{20}{l}} {\frac{1}{n}}&{{\text{for }}1 \leqslant k \leqslant n} \\ 0&{{\text{otherwise}}} \end{array}} \right.\)       AG

[4 marks]

\({\text{E}}({X_n}) = 0 \times 0 + 1 \times \frac{1}{n} + 2 \times \frac{1}{n} + 3 \times \frac{1}{n} +  \ldots n \times \frac{1}{n} + (n + 1) \times 0 +  \ldots  \times 0\)     (M1)(A1)

\( = \frac{1}{n} \times \sum\limits_{k = 1}^{k = n} k \)

\( = \frac{1}{n} \times \frac{1}{2}n(n + 1) = \frac{{n + 1}}{2}\)    A1

Note:     Accept use of \(G'(1)\).

[3 marks]

\({X_{n - 1}}\) and \({X_{n + 1}}\) are independent \( \Rightarrow {\text{E}}({X_{n - 1}} \times {X_{n + 1}}) = {\text{E}}({X_{n - 1}}) \times {\text{E}}({X_{n + 1}})\)     M1

\( = \frac{n}{2} \times \frac{{n + 2}}{2}\)  A1

required to solve \({n^2} < 6n{\text{ }}({\text{or }}n + 2 < 8)\)     M1

solution: \((2 \leqslant ){\text{ }}n < 6\)     A1

[4 marks]

Again this was a question that tested candidates and although many started only a very limited number were able to make significant progress. Part (a) was rarely done well with most candidates not understanding what was required. There was a little more success with part (b) but a number of candidates attempted methods that were not going to lead to anything meaningful. Most candidates did not understand what was required from part (c) and few correct answers were seen, even taking into account the fact that follow through marks could be awarded from (b).

Again this was a question that tested candidates and although many started only a very limited number were able to make significant progress. Part (a) was rarely done well with most candidates not understanding what was required. There was a little more success with part (b) but a number of candidates attempted methods that were not going to lead to anything meaningful. Most candidates did not understand what was required from part (c) and few correct answers were seen, even taking into account the fact that follow through marks could be awarded from (b).

Again this was a question that tested candidates and although many started only a very limited number were able to make significant progress. Part (a) was rarely done well with most candidates not understanding what was required. There was a little more success with part (b) but a number of candidates attempted methods that were not going to lead to anything meaningful. Most candidates did not understand what was required from part (c) and few correct answers were seen, even taking into account the fact that follow through marks could be awarded from (b).