Edexcel IAL October 2023 P4 (WMA14/01) Q1 Binomial Expansion
(a) Find the first four terms, in ascending powers of x, of the binomial expansion of
\[\frac{8}{(2-5x)^2}\]
writing each term in simplest form.(4)
(b) Find the range of values of x for which this expansion is valid.(1)
Solution to Part a: Find the first four terms of \(\frac{8}{(2-5x)^2}\)
Key Concepts Used
- Algebraic Preparation: Factor out 2.
- General Binomial Expansion Formula for \(n = -2\)
Step-by-Step Solution
- Prepare the expression
Write in the index form.
\[\frac{8}{(2-5x)^2} = 8(2-5x)^{-2}\]
Let’s factor out 2.
\[= 8\left[2\left(1-\frac{5}{2}x\right)\right]^{-2}\]
\[= 8 \times 2^{-2}\left(1-\frac{5}{2}x\right)^{-2}\]
\[= 8 \times \frac{1}{4}\left(1-\frac{5}{2}x\right)^{-2}\]
So,
\[\frac{8}{(2-5x)^2} = 2\left(1+\left(-\frac{5}{2}x\right)\right)^{-2}\] - Identify \(n\) and \(X\):
\(n = -2\) and \(X = -\frac{5}{2}x\).
- Apply the Binomial Formula
\[(1 + x)^n = x^0 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots\]
Term in \(x^0\):
– 1Term in \(x^1\):
– \(nX = (-2)\left(-\frac{5}{2}x\right) = 5x\)Term in \(x^2\):
– \(\frac{n(n-1)}{2!}X^2 = \frac{(-2)(-3)}{2}\left(-\frac{5}{2}x\right)^2 = 3\left(\frac{25}{4}x^2\right) = \frac{75}{4}x^2\)Term in \(x^3\):
– \(\frac{n(n-1)(n-2)}{3!}X^3 = \frac{(-2)(-3)(-4)}{6}\left(-\frac{5}{2}x\right)^3 = -4\left(\frac{125}{8}x^3\right) = -\frac{125}{2}x^3\) - Multiply by 2:
\[2\left(1 + 5x + \frac{75}{4}x^2 – \frac{125}{2}x^3 + \ldots\right) = 2 + 10x + \frac{75}{2}x^2 – 125x^3 + \ldots\]
Final Answer
\[\frac{8}{\left(2 – 5x\right)^2} = 2 + 10x + \frac{75}{2}x^2 – 125x^3 + \ldots\]
Solution to Part b: Find the range of values of \(x\) for which this expansion is valid.
Key Concepts Used
- Range of Validity: The expansion is valid when \(|X| < 1\), where \(X = -\frac{5}{2}x\).
Step-by-Step Solution
- Set up the inequality:
Valid for \(|-\frac{5}{2}x| < 1\).
- Solve for \(x\):
\(|\frac{5}{2}x| < 1 \implies |x| < \frac{2}{5}\).
Final Answer (Part b)
\[|x| < \frac{2}{5}\]