19N.2.AHL.TZ0.H_5

pestleMathematicsAAHLPaper 219N· sl-1-9-binomial-theorem-where-n-is-an-integersource ↗

Consider the expansion of ${\left( {2 + x} \right)^n}$ , where $n \geqslant 3$ and $n \in \mathbb{Z}$ .

The coefficient of ${x^3}$ is four times the coefficient of ${x^2}$ . Find the value of $n$ .

Markscheme / solution

attempt to find coefficients in binomial expansion (M1)

coefficient of ${x^2}\,{\text{:}}\,\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right) \times {2^{n - 2}}$ ; coefficient of ${x^3}\,{\text{:}}\,\left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right) \times {2^{n - 3}}$ A1A1

Note: Condone terms given rather than coefficients. Terms may be seen in an equation such as that below.

$\left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right) \times {2^{n - 3}} = 4\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right) \times {2^{n - 2}}$ (A1)

attempt to solve equation using GDC or algebraically (M1)

$\left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right) = 8\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right)$

$\frac{{n{\text{!}}}}{{3{\text{!}}\left( {n - 3} \right){\text{!}}}} = \frac{{8n{\text{!}}}}{{2{\text{!}}\left( {n - 2} \right){\text{!}}}}$

$\frac{1}{3} = \frac{8}{{n - 2}}$

$n = 26$ A1

[6 marks]

Examiners’ report

[N/A]