Consider the expansion of ${x^2}{\left( {3{x^2} + \frac{k}{x}} \right)^8}$. The constant term is ${\text{16 128}}$.

Find $k$.

valid approach     (M1)

eg     ${\text{\(\left( \begin{array}{c}8\\r\end{array} \right)\)}}{\left( {3{x^2}} \right)^{8 - r}}{\left( {\frac{k}{x}} \right)^r}$,

${\left( {3{x^2}} \right)^8} + {\text{\(\left( \begin{array}{c}8\\1\end{array} \right)\)}}{\left( {3{x^2}} \right)^7}\left( {\frac{k}{x}} \right) + {\text{\(\left( \begin{array}{c}8\\2\end{array} \right)\)}}{\left( {3{x^2}} \right)^6}{\left( {\frac{k}{x}} \right)^2} +  \ldots$, Pascal’s triangle to ${9^{{\text{th}}}}$ line

attempt to find value of r which gives term in ${x^0}$     (M1)

eg     exponent in binomial must give ${x^{ - 2}},{\text{ }}{x^2}{\left( {{x^2}} \right)^{8 - r}}{\left( {\frac{k}{x}} \right)^r} = {x^0}$

correct working     (A1)

eg     $2(8 - r) - r =  - 2,{\text{ }}18 - 3r = 0,{\text{ }}2r + ( - 8 + r) =  - 2$

evidence of correct term     (A1)

eg     ${\text{\(\left( \begin{array}{c}8\\2\end{array} \right)\), \(\left( \begin{array}{c}8\\6\end{array} \right)\)}}{\left( {3{x^2}} \right)^2}{\left( {\frac{k}{x}} \right)^2},{\text{ }}r = 6,{\text{ }}r = 2$

equating their term and 16128 to solve for $k$     M1

eg     ${x^2}{\text{\(\left( \begin{array}{c}8\\6\end{array} \right)\)}}{\left( {3{x^2}} \right)^2}{\left( {\frac{k}{x}} \right)^6} = 16128,{\text{ }}{k^6} = \frac{{16128}}{{28(9)}}$

$k =  \pm 2$     A1A1     N2

Note: If no working shown, award N0 for $k = 2$.

Total [7 marks]