Edexcel IAL June 2022 P4 (WMA14/01) Q1 Binomial Expansion
The binomial expansion of
\[(3 + kx)^{-2} \quad |kx| < 3\]
where \( k \) is a non-zero constant, may be written in the form
\[A + Bx + Cx^2 + Dx^3 + \ldots\]
where \( A, B, C \) and \( D \) are constants.
(a) Find the value of \( A \)
(1)
Given that \( C = 3B \)
(b) show that \( k^2 + 6k = 0 \)
(c) Hence
(i) Find the value of \( k \)
(ii) Find the value of \( D \)
(3)
Solution to Part a: Find the value of \( A \).
Key Concepts Used
- Algebraic Preparation: Factor out 3.
- Constant Term: \( A \) is the constant term in the expansion.
Step-by-Step Solution
- Prepare the expression
\[(3 + kx)^{-2} = \left[ 3 \left( 1 + \frac{kx}{3} \right) \right]^{-2} = 3^{-2} \left( 1 + \frac{kx}{3} \right)^{-2} = \frac{1}{9} \left( 1 + \frac{kx}{3} \right)^{-2}\]
- Find \( A \)
\( A \) is the constant factor \(\frac{1}{9}\) multiplied by the constant term (1) of the bracket expansion.
\[A = \frac{1}{9}\]
Final Answer
\[A = \frac{1}{9}\]
Solution to Part b: Show that \( k^2 + 6k = 0 \) given \( C = 3B \).
Key Concepts Used
- Coefficient Calculation: Finding \( B \) (for \( x^1 \)) and \( C \) (for \( x^2 \)).
- Algebraic Derivation: Substituting \( B \) and \( C \) into the given condition.
Step-by-Step Solution
- Calculate \( B \) (Coefficient of \( x^1 \))
\( B \) is \( \frac{1}{9} \times (nX) \).
\[ B = \frac{1}{9} \times (-2) \left( \frac{k}{3} \right) = -\frac{2k}{27} \] - Calculate \( C \) (Coefficient of \( x^2 \))
\( C \) is \( \frac{1}{9} \times \left( \frac{n(n-1)}{2!} X^2 \right) \).
\[ C = \frac{1}{9} \times \left( \frac{(-2)(-3)}{2} \right) \left( \frac{k}{3} \right)^2 = \frac{1}{9} \times (3) \left( \frac{k^2}{9} \right) = \frac{3k^2}{81} = \frac{k^2}{27} \] - Use the condition \( C = 3B \)
\[ \frac{k^2}{27} = 3 \times \left( -\frac{2k}{27} \right) \]
- Derive the result
Multiply both sides by 27 to cancel out denominators.
\[ k^2 = -6k \]
\[ k^2 + 6k = 0 \]
Final Answer
\[ k^2 + 6k = 0 \]
Solution to Part c(i): Find the value of k.
Key Concepts Used
- Solving Quadratic: Factorisation.
- Contextual Constraint: \( k \) is a non-zero constant.
Step-by-Step Solution
- Solve the equation
\[k^2 + 6k = 0 \Rightarrow k(k + 6) = 0\]
- Possible value
\[k = 0 \text{ or } k = -6\]
Apply constraint: Since \( k \) is a non-zero constant, \( k \) must be \(-6\).
Final Answer
\[k = -6\]
Solution to Part c(ii): Find the value of D.
Key Concepts Used
- Coefficient Calculation: Finding \( D \) (coefficient of \( x^3 \)).
- Substitution: Use \( k = -6 \).
Step-by-Step Solution
- Calculate \( D \) (Coefficient of \( x^3 \))
\( D \) is \( \frac{1}{9} \times \left( \frac{n(n-1)(n-2)}{3!} X^3 \right) \).
\[ D = \frac{1}{9} \times \left( \frac{(-2)(-3)(-4)}{6} \right) \left( \frac{k}{3} \right)^3 = \frac{1}{9} \times (-4) \left( \frac{k^3}{27} \right) = -\frac{4k^3}{243} \] - Substitute \( k = -6 \):
\[ D = -\frac{4(-6)^3}{243} = -\frac{4(-216)}{243} = \frac{864}{243} \]
- Simplify
Divide numerator and denominator by 81 (or 27 twice).
\[ D = \frac{32}{9} \]
Final Answer
\[D = \frac{32}{9}\]