21M.2.SL.TZ1.6

pestleMathematicsAASLPaper 221M· sl-1-9-binomial-theorem-where-n-is-an-integersource ↗

Consider the expansion of $(3 + x^{2})^{n + 1}$ , where $n \in ℤ^{+}$ .

Given that the coefficient of $x^{4}$ is $20 412$ , find the value of $n$ .

Markscheme / solution

METHOD 1

product of a binomial coefficient, a power of $3$ (and a power of $x^{2}$ ) seen (M1)

evidence of correct term chosen (A1)

${}^{{}_{n + 1}}C_{2} \times 3^{n + 1 - 2} \times {(x^{2})}^{2} (= \frac{n (n + 1)}{2} \times 3^{n - 1} \times x^{4})$ OR $n - r = 1$

equating their coefficient to $20412$ or their term to $20412 x^{4}$ (M1)

EITHER

${}^{{}_{n + 1}}C_{2} \times 3^{n - 1} = 20412$ (A1)

${}^{r + 2}C_{r} \times 3^{r} = 20412 \Rightarrow r = 6$ (A1)

THEN

$n = 7$ A1

METHOD 2

$3^{n + 1} {(1 + \frac{x^{2}}{3})}^{n + 1}$

product of a binomial coefficient, and a power of $\frac{x^{2}}{3}$ OR $\frac{1}{3}$ seen (M1)

evidence of correct term chosen (A1)

$3^{n + 1} \times \frac{n (n + 1)}{2!} \times {(\frac{x^{2}}{3})}^{2} (= \frac{3^{n - 1}}{2} n (n + 1) x^{4})$

equating their coefficient to $20412$ or their term to $20412 x^{4}$ (M1)

$3^{n - 1} \times \frac{n (n + 1)}{2} = 20412$ (A1)

$n = 7$ A1

[5 marks]

Examiners’ report

[N/A]