Edexcel IAL January 2022 P4 (WMA14/01) Q2 Binomial Expansion & Approximation
(a) Find, in ascending powers of \( x \), the first three non-zero terms of the binomial series expansion of
\[\sqrt[3]{1 + 4x^3} \quad |x| < \frac{1}{\sqrt[3]{4}}\]
giving each coefficient as a simplified fraction.
(4)
(b) Use the expansion from part (a) with \( x = \frac{1}{3} \) to find a rational approximation to \(\sqrt[3]{31}\).
(3)
Solution to Part a: Find the first three non-zero terms of \( \sqrt[3]{1 + 4x^3} \).
Key Concepts Used
- General Binomial Expansion for \( n = 1/3 \).
- Substitution: Note that \( X = 4x^3 \).
Step-by-Step Solution
-
Identify \( n \) and \( X \):
\( n = \frac{1}{3} \) and \( X = 4x^3 \).
-
Apply the Binomial Formula
– Term in \( x^0 \):
\[ X^0 = 1 \]
– Term in \( x^3 \):
\[ nX = \frac{1}{3}(4x^3) = \frac{4}{3}x^3 \]
– Term in \( x^6 \):
\[ \frac{n(n-1)}{2!}X^2 = \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2}(4x^3)^2 = -\frac{1}{9}(16x^6) = -\frac{16}{9}x^6 \]
Final Answer
\[1 + \frac{4}{3}x^3 – \frac{16}{9}x^6 + …\]
Solution to Part b: Find a rational approximation to \(\sqrt[3]{31}\) using \( x = \frac{1}{3} \).
Key Concepts Used
-
Relate function to \(\sqrt[3]{31}\)
Substitute \(x = \frac{1}{3}\) into the function, \(\sqrt[3]{1 + 4x^3}\):
\[\sqrt[3]{1 + 4\left(\frac{1}{3}\right)^3} = \sqrt[3]{1 + \frac{4}{27}} = \sqrt[3]{\frac{31}{27}} = \frac{\sqrt[3]{31}}{3}\]
Thus, the expansion approximates, \(\frac{\sqrt[3]{31}}{3}\). -
Calculate the approximation for \(\frac{\sqrt[3]{31}}{3}\)
Substitute \(x = \frac{1}{3}\) into the expansion found in part a:
\[\frac{\sqrt[3]{31}}{3} \approx 1 + \frac{4}{3}\left(\frac{1}{3}\right)^3 – \frac{16}{9}\left(\frac{1}{3}\right)^6\]
\[\frac{\sqrt[3]{31}}{3} \approx 1 + \frac{4}{81} – \frac{16}{6561}\] -
Multiply by 3 on both sides and simplify:
\[\sqrt[3]{31} \approx 3\left(1 + \frac{4}{81} – \frac{16}{6561}\right)\]
\[\sqrt[3]{31} \approx 3 + \frac{4}{27} – \frac{16}{2187} = \frac{6869}{2187}\]
Final Answer
\[\sqrt[3]{31} \approx \frac{6869}{2187}\]