Given that \({\left( {1 + \frac{2}{3}x} \right)^n}{(3 + nx)^2} = 9 + 84x + \ldots \) , find the value of n .

attempt to expand \({\left( {1 + \frac{2}{3}x} \right)^n}\)     (M1)

e.g. Pascal's triangle, \({\left( {1 + \frac{2}{3}x} \right)^n} = 1 + \frac{2}{3}nx + \ldots \)

correct first two terms of \({\left( {1 + \frac{2}{3}x} \right)^n}\) (seen anywhere)     (A1)

e.g. \(1 + \frac{2}{3}nx\)

correct first two terms of quadratic (seen anywhere)     (A1)

e.g. 9 , \(6nx\) , \((9 + 6nx + {n^2}{x^2})\)

correct calculation for the x-term     A2

e.g. \(\frac{2}{3}nx \times 9 + 6nx\) , \(6n + 6n\) , \(12n\)

correct equation     A1

e.g. \(6n + 6n = 84\) , \(12nx = 84x\)

\(n = 7\)     A1     N1

[7 marks]

This question proved quite challenging for the majority of candidates, although there were a small number who were able to find the correct value of n using algebraic and investigative methods. While most candidates recognized the need to apply the binomial theorem, the majority seemed to have no idea how to do so when the exponent was a variable, n, rather than a known integer. Most candidates who attempted this question did expand the quadratic correctly, but many went no further, or simply set the x-term of the quadratic equal to \(84x\), ignoring the expansion of the first binomial altogether.