Expand $\left(2x – \dfrac{1}{x}\right)^{4}$ fully, and state the term that does not depend on $x$.
Question HKC78-AAHL-BIN-01
[Maximum mark: 5]
Expand $\left(2x – \dfrac{1}{x}\right)^{4}$ fully, and state the term that does not depend on $x$.
Mark Scheme
The general term is $dbinom{4}{k}(2x)^{4-k}left(-dfrac{1}{x}right)^{k}$.
$k=0$: $16x^{4}$. $k=1$: $dbinom{4}{1}(8x^{3})left(-dfrac{1}{x}right) = -32x^{2}$. $k=2$: $dbinom{4}{2}(4x^{2})left(dfrac{1}{x^{2}}right) = 24$. $k=3$: $dbinom{4}{3}(2x)left(-dfrac{1}{x^{3}}right) = -dfrac{8}{x^{2}}$. $k=4$: $dfrac{1}{x^{4}}$. [4]
So $left(2x – dfrac{1}{x}right)^{4} = 16x^{4} – 32x^{2} + 24 – dfrac{8}{x^{2}} + dfrac{1}{x^{4}}$. The term independent of $x$ is $24$. [1]
Total: 5 marks
In the expansion of $(1 + 3x)^{n}$, where $n$ is an integer and $n > 2$, the coefficient of $x^{2}$ is $135$. Find $n$.
Question HKC78-AAHL-BIN-02
[Maximum mark: 5]
In the expansion of $(1 + 3x)^{n}$, where $n$ is an integer and $n > 2$, the coefficient of $x^{2}$ is $135$. Find $n$.
Mark Scheme
The coefficient of $x^{2}$ is $dbinom{n}{2} cdot 3^{2} = 135$. [1]
$dfrac{n(n-1)}{2} cdot 9 = 135$, so $n(n-1) = 30$. [2]
$n^{2} – n – 30 = 0$, hence $(n-6)(n+5) = 0$. [1]
$n = 6$, because $n$ is an integer greater than $2$. [1]
Total: 5 marks
