Binomial Expansion

Coach Name: Sir Muhammad Abdullah Shah

Edexcel IAL January 2024 P4 (WMAA14/01) Q1 Binomial Expansion

Find, in ascending powers of \( x \) up to and including the term in \( x^3 \), the binomial expansion of

\[(1 – 4x)^{-3} \quad |x| < \frac{1}{4}\]

fully simplifying each term.(4)

Solution to Question: Expansion of \((1 – 4x)^{-3}\)
Key Concepts Used
  1. General Binomial Expansion Formula for \(n = -3\).
Step-by-Step Solution
  1. Identify \(n\) and \(X\)
    The expression is already in \((1 + X)^n\) form. \(n = -3\) and \(X = -4x\).
  2. Apply the Binomial Formula:
    \[ (1 + x)^n = x^0 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + …\]

    Term in \(x^0\):
    \[\begin{array}{ccc}
    1 & & \\
    \text{Term in } x^1: & nX = (-3)(-4x) = 12x
    \end{array}\]
    Term in \(x^2\):
    \[\frac{n(n-1)}{2!} x^2 = \frac{(-3)(-4)}{2} (-4x)^2 = 6(16x^2) = 96x^2\]
    Term in \(x^3\):
    \[\frac{n(n-1)(n-2)}{3!} x^3 = \frac{(-3)(-4)(-5)}{6} (-4x)^3 = -10(-64x^3) = 640x^3\]

Final Answer
\[\left[ (1 – 4x)^{-3} = 1 + 12x + 96x^2 + 640x^3 + … \right]\]