In the expansion of $a{x^3}{(2 + ax)^{11}}$, the coefficient of the term in ${x^5}$ is 11880. Find the value of $a$.

valid approach for expansion (must have correct substitution for parameters, but accept an incorrect value for $r$)     (M1)

eg$\,\,\,\,\,$$\left( {\begin{array}{*{20}{c}} {11} \\ r \end{array}} \right){(2)^{11 - r}}a{x^r},{\text{ }}\left( {\begin{array}{*{20}{c}} {11} \\ 3 \end{array}} \right){(2)^8}{(ax)^3},{\text{ }}{2^{11}} + \left( {\begin{array}{*{20}{c}} {11} \\ 1 \end{array}} \right){(2)^{10}}{(ax)^1} + \left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){(2)^9}(ax) + \ldots$

recognizing need to find term in ${x^2}$ in binomial expansion     (A1)

eg$\,\,\,\,\,$$r = 2,{\text{ }}{(ax)^2}$

correct term or coefficient in binomial expansion (may be seen in equation)     (A1)

eg$\,\,\,\,\,$$\left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){(ax)^2}{(2)^9},{\text{ }}55({a^2}{x^2})(512),{\text{ }}28160{a^2}$

setting up equation in ${x^5}$ with their coefficient/term (do not accept other powers of $x$)     (M1)

eg$\,\,\,\,\,$$a{x^3}\left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){(ax)^2}{(2)^9} = 11880{x^5}$

correct equation     (A1)

eg$\,\,\,\,\,$$28160{a^3} = 11880$

$a = \frac{3}{4}$     A1     N3

[6 marks]