Write down the first four terms of the expansion.

(i)     Show that ${n^3} - 9{n^2} + 14n = 0$.

(ii)     Hence find the value of $n$.

$1,{\text{ }}nx,{\text{ }}\frac{{n(n - 1)}}{2}{x^2},{\text{ }}\frac{{n(n - 1)(n - 2)}}{6}{x^3}$    A1A1

Note:     Award A1 for the first two terms and A1 for the next two terms.

Note:     Accept $\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right)$ notation.

Note:     Allow the terms seen in the context of an arithmetic sum.

Note:     Allow unsimplified terms, eg, those including powers of 1 if seen.

[2 marks]

(i)     EITHER

using ${u_3} - {u_2} = {u_4} - {u_3}$     (M1)

$\frac{{n(n - 1)}}{2} - n = \frac{{n(n - 1)(n - 2)}}{6} - \frac{{n(n - 1)}}{2}$    A1

attempting to remove denominators and expanding (or vice versa)     M1

$3{n^2} - 9n = {n^3} - 6{n^2} + 5n$ (or equivalent, eg, $6{n^2} - 12n = {n^3} - 3{n^2} + 2n$)     A1

OR

using ${u_2} + {u_4} = 2{u_3}$     (M1)

$n + \frac{{n(n - 1)(n - 2)}}{6} = n(n - 1)$    (A1)

attempting to remove denominators and expanding (or vice versa)     M1

$6n + {n^3} - 3{n^2} + 2n = 6{n^2} - 6n$ (or equivalent)     (A1)

THEN

${n^3} - 9{n^2} + 14n = 0$    AG

(ii)     $n(n - 2)(n - 7) = 0$ or $(n - 2)(n - 7) = 0$     (A1)

$n = 7$ only (as $n \geqslant 3$)     A1

[6 marks]

This was another question that was very well answered by most candidates.

This was another question that was very well answered by most candidates. Some struggled in part (b) by attempting to find an expression for $d$, a common difference, then substituting this in to further equations, where algebra tended to falter. The most fruitful technique was to apply ${u_3} - {u_2} = {u_4} - {u_3}$. Good presentation often helped candidates reach the final result. Correct factorisation was more often seen than not in the final section, though a small number thought it judicious to guess the correct answer(s) here.