Edexcel IAL June 2023 P4 (WMA14/01) Q1 Binomial Expansion
(a) Find the first 4 terms of the binomial expansion, in ascending powers of \(x\), of
\[\left( \frac{1}{4} – \frac{1}{2}x \right)^{-\frac{3}{2}} \quad |x| < \frac{1}{2}\]
giving each term in simplest form.
Given that
\[\left( \frac{1}{4} – \frac{1}{2}x \right)^n \left( \frac{1}{4} – \frac{1}{2}x \right)^{-\frac{3}{2}} = \left( \frac{1}{4} – \frac{1}{2}x \right)^{\frac{1}{2}}\]
(b) write down the value of \(n\).(1)
(c) Hence, or otherwise, find the first 3 terms of the binomial expansion, in ascending powers of \(x\), of
\[\left( \frac{1}{4} – \frac{1}{2}x \right)^{\frac{1}{2}} \quad |x| < \frac{1}{2}\]
giving each term in simplest form.(3)
Solution to Part a: Expansion of \(\left( \frac{1}{4} – \frac{1}{2}x \right)^{-\frac{3}{2}}\)
Key Concepts Used
- Algebraic Preparation: Factor out 1/4.
- General Binomial Expansion Formula for n=1/2.
Step-by-Step Solution
- Prepare the expression
Factor out \(\frac{1}{4}\) and rewrite the root as a power:
\[\left( \frac{1}{4} – \frac{1}{2}x \right)^{-\frac{3}{2}} = \left( \frac{1}{4}(1 – 2x) \right)^{-\frac{3}{2}} = \left( \frac{1}{4} \right)^{-\frac{3}{2}}(1 – 2x)^{-\frac{3}{2}} = 8(1 – 2x)^{-\frac{3}{2}}\] - Identify \(n\) and \(X\)
\[ n = -\frac{3}{2} \text{ and } X = -2x \]
- Apply the Binomial Formula:
– Term in \( x^0 \):
\[x^0 = 1\]
– Term in \( x^1 \):
\[nX = \left( -\frac{3}{2} \right) (-2x) = 3x\]
– Term in \( x^2 \):
\[\frac{n(n-1)}{2!} X^2 = \frac{\left( -\frac{3}{2} \right) \left( -\frac{3}{2} \right)}{2!} (-2x)^2 = \frac{15}{2} x^2\]
– Term in \( x^3 \):
\[\frac{n(n-1)(n-2)}{3!} X^3 = \frac{\left( -\frac{3}{2} \right) \left( -\frac{3}{2} \right)}{3!} (-2x)^3 = \frac{35}{2} x^3\] - Multiply by 8
\[8(1 – 2x)^{\frac{3}{2}} = 8 \left[ 1 + 3x + \frac{15}{2} x^2 + \frac{35}{2} x^3 + \cdots \right]\]
\[= 8 + 24x + 60x^2 + 140x^3 + \cdots\]
Final Answer
\[\left( \frac{1}{4} – \frac{1}{2} x \right)^{-\frac{3}{2}} = 8 + 24x + 60x^2 + 140x^3 + \cdots\]
Solution to Part b: Write down the value of \( n \) given \( \left( \frac{1}{4} – \frac{1}{2}x \right)^n \left( \frac{1}{4} – \frac{1}{2}x \right)^{-\frac{n}{2}} = \left( \frac{1}{4} – \frac{1}{2}x \right)^{\frac{1}{2}} \)
Key Concepts Used
- Laws of indices: When multiplying terms with the same base, powers are added.
\[ a^m a^n = a^{m+n} \]
Step-by-Step Solution
- Set up the index equation
For our easiness let
\[ P = \frac{1}{4} – \frac{1}{2}x \]
The given relationship is \( P^n P^{-\frac{3}{2}} = P^{\frac{1}{2}} \) - Solve for \( n \)
Equate the exponents:
\[ n – \frac{3}{2} = \frac{1}{2} \]
\[ n = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2 \]
Final Answer
\[ n = 2 \]
Solution to Part c: Find the first 3 terms of the expansion of \(\left(\frac{1}{4} – \frac{1}{2}x\right)^{\frac{1}{2}}\)
Key Concepts Used
- Combining Expansions: Multiplying the expansion found in part (a) by the expansion of \(\left(\frac{1}{4} – \frac{1}{2}x\right)^n\)
Step-by-Step Solution
- Expand \(\left(\frac{1}{4} – \frac{1}{2}x\right)^2\)
\[\left(\frac{1}{4} – \frac{1}{2}x\right)^2 = \left(\frac{1}{4}\right)^2 + 2\left(\frac{1}{4}\right)\left(-\frac{1}{2}x\right) + \left(-\frac{1}{2}x\right)^2\]
\[= \frac{1}{16} – \frac{1}{4}x + \frac{1}{4}x^2\] - Use the given relationship
Since we know that from the previous part,
\[\left(\frac{1}{4} – \frac{1}{2}x\right)^2 \left(\frac{1}{4} – \frac{1}{2}x\right)^{-\frac{3}{2}} = \left(\frac{1}{4} – \frac{1}{2}x\right)^{\frac{1}{2}}\]
So, to find the expansion of \(\left(\frac{1}{4} – \frac{1}{2}x\right)^{\frac{1}{2}}\), let’s simply use the expansion of \(\left(\frac{1}{4} – \frac{1}{2}x\right)^2\) & \(\left(\frac{1}{4} – \frac{1}{2}x\right)^{-\frac{3}{2}}\) found previously.
\[\left(\frac{1}{4} – \frac{1}{2}x\right)^{\frac{1}{2}} = \left(\frac{1}{16} – \frac{1}{4}x + \frac{1}{4}x^2\right)(8 + 24x + 60x^2 + \cdots)\] - Expanding the brackets
\[\left(\frac{1}{4} – \frac{1}{2}x\right)^{\frac{1}{2}} = \frac{8}{16} + \frac{24x}{16} + \frac{60x^2}{16} + \frac{8x}{4} + \frac{24x^2}{4}\]
\[\left(\frac{1}{4} – \frac{1}{2}x\right)^{\frac{1}{2}} = \frac{1}{2} + \frac{3}{2}x + \frac{15x^2}{4} – 2x – 6x^2 + 2x^2 + \cdots\]
\[= \frac{1}{2} – \frac{1}{2}x – \frac{1}{4}x^2 + \cdots\]
Final Answer
\[\left(\frac{1}{4} – \frac{1}{2}x\right)^{\frac{1}{2}} = \frac{1}{2} – \frac{1}{2}x – \frac{1}{4}x^2 + \cdots\]