In the expansion of ${(3x + 1)^n}$, the coefficient of the term in ${x^2}$ is $135n$, where $n \in {\mathbb{Z}^ + }$. Find $n$.

Note:     Accept sloppy notation (such as missing brackets, or binomial coefficient which includes ${x^2}$).

evidence of valid binomial expansion with binomial coefficients     (M1)

eg$\;\;\;\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){(3x)^r}{(1)^{n - r}},{\text{ }}{(3x)^n} + n{(3x)^{n - 1}} + \left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){(3x)^{n - 2}} +  \ldots ,{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){(1)^{n - r}}{(3x)^r}$

attempt to identify correct term     (M1)

eg$\;\;\;\left( {\begin{array}{*{20}{c}} n \\ {n - 2} \end{array}} \right),{\text{ }}{(3x)^2},{\text{ }}n - r = 2$

setting correct coefficient or term equal to $135n$ (may be seen later)     A1

eg$\;\;\;9\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right) = 135n,{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ {n - 2} \end{array}} \right){(3x)^2} = 135n,{\text{ }}\frac{{9n(n - 1)}}{2} = 135n{x^2}$

correct working for binomial coefficient (using $_n{C_r}$ formula)     (A1)

eg$\;\;\;\frac{{n(n - 1)(n - 2)(n - 3) \ldots }}{{2 \times 1 \times (n - 2)(n - 3)(n - 4) \ldots }},{\text{ }}\frac{{n(n - 1)}}{2}$

EITHER

evidence of correct working (with linear equation in $n$)     (A1)

eg$\;\;\;\frac{{9(n - 1)}}{2} = 135,{\text{ }}\frac{{9(n - 1)}}{2}{x^2} = 135{x^2}$

correct simplification     (A1)

eg$\;\;\;n - 1 = \frac{{135 \times 2}}{9},{\text{ }}\frac{{(n - 1)}}{2} = 15$

$n = 31$     A1     N2

OR

evidence of correct working (with quadratic equation in $n$)     (A1)

eg$\;\;\;9{n^2} - 279n = 0,{\text{ }}{n^2} - n = 30n,{\text{ (9}}{{\text{n}}^2} - 9n){x^2} = 270n{x^2}$

evidence of solving     (A1)

eg$\;\;\;9n(n - 31) = 0,{\text{ }}9{n^2} = 279n$

$n = 31$     A1     N2

Note:     Award A0 for additional answers.

[7 marks]