Question
The binomial expansion of \( (1 + kx)^n \) is given by \( 1 + 12x + 28k^2x^2 + \ldots + k^nx^n \), where \( n \in \mathbb{Z}^+ \) and \( k \in \mathbb{Q} \).
Find the value of \( n \) and the value of \( k \).
▶️Answer/Explanation
Detail Solution
Step 1: Identify the coefficients in the binomial expansion.
The binomial expansion of \( (1 + kx)^n \) is given by the formula:
\[
\sum_{r=0}^{n} \binom{n}{r} (kx)^r = \sum_{r=0}^{n} \binom{n}{r} k^r x^r
\]
From the given expansion \( 1 + 12x + 28k^2x^2 + \ldots + k^nx^n \), we can identify the coefficients for \( x^1 \) and \( x^2 \).
Step 2: Set up equations based on the coefficients.
The coefficient of \( x^1 \) is \( \binom{n}{1} k = 12 ……..(1)\).
The coefficient of \( x^2 \) is \( \binom{n}{2} k^2 = 28\times k^2………(2) \).
Step 3: Express \( k \) in terms of \( n \) from the first equation.
From \( \binom{n}{1} k = 12 \), we have:
\[
n k = 12
\]
Step 4: from the second equation.
Calculating \( \binom{n}{2} = \frac{n(n-1)}{2} \), we get:
\[
\frac{n(n-1)}{2} \times k^2 = 28\times k^2
\]
$$ \frac{n(n-1)}{2}=28$$
This simplifies to:
\[
n\times (n-1)=56
\]
Step 5: Clear the fraction by multiplying both sides by \( n \).
\[
n^2 -n -56=0
\]
$$n^2 -8n +7n -56=0$$
$$n(n-8) +7(n-8)=0$$
$$(n-8)(n+7)=0$$
$$(n-8)=0 \Rightarrow n=0,(n+7)=0 \Rightarrow n=-7$$(which is not possible)
from euation first $$nk=12 \Rightarrow 8\times k=12 \Rightarrow k= \frac{3}{2}$$
$$ n=8, k=\frac{3}{2}$$
————Markscheme—————–
From the binomial expansion:
\( nk = 12 \) and \( \frac{n(n-1)}{2}k^2 = 28 \)
Solving these equations, we get \( n = 8 \) and \( k = \frac{3}{2} \).
Question
Expand and simplify \({\left( {{x^2} – \frac{2}{x}} \right)^4}\).
▶️Answer/Explanation
\({\left( {{x^2} – \frac{2}{x}} \right)^4} = {({x^2})^4} + 4{({x^2})^3}\left( { – \frac{2}{x}} \right) + 6{({x^2})^2}{\left( { – \frac{2}{x}} \right)^2} + 4({x^2}){\left( { – \frac{2}{x}} \right)^3} + {\left( { – \frac{2}{x}} \right)^4}\) (M1)
\( = {x^8} – 8{x^5} + 24{x^2} – \frac{{32}}{x} + \frac{{16}}{{{x^4}}}\) A3
Note: Deduct one A mark for each incorrect or omitted term. [4 marks]
Question
Expand \({(2 – 3x)^5}\) in ascending powers of x, simplifying coefficients.
▶️Answer/Explanation
clear attempt at binomial expansion for exponent 5 M1
\({2^5} + 5 \times {2^4} \times ( – 3x) + \frac{{5 \times 4}}{2} \times {2^3} \times {( – 3x)^2} + \frac{{5 \times 4 \times 3}}{6} \times {2^2} \times {( – 3x)^3}\)
\( + \frac{{5 \times 4 \times 3 \times 2}}{{24}} \times 2 \times {( – 3x)^4} + {( – 3x)^5}\) (A1)
Note: Only award M1 if binomial coefficients are seen.
\( = 32 – 240x + 720{x^2} – 1080{x^3} + 810{x^4} – 243{x^5}\) A2
Note: Award A1 for correct moduli of coefficients and powers. A1 for correct signs.
Total [4 marks]