Edexcel IAL January 2021 P4 (WMA14/01) Q1 Binomial Expansion & Approximation
(a) Find the first 4 terms, in ascending powers of \(x\), of the binomial expansion of
\[\left(\frac{1}{4} – 5x\right)^{\frac{1}{2}} \quad |x| < \frac{1}{20}\]
giving each coefficient in its simplest form.
(5)
By substituting \(x = \frac{1}{100}\) into the answer for (a)
(b) Find an approximation for \(\sqrt{5}\).
(c) Give your answer in the form \(\frac{a}{b}\) where a and b are integers to be found.
(2)
Solution to Part a: Find the first 4 terms of \(\left(\frac{1}{4} – 5x\right)^{\frac{1}{2}}\).
Key Concepts Used
- Algebraic Preparation: Factor out \(\frac{1}{4}\).
- General Binomial Expansion Formula for \(n = 1/2\).
Step-by-Step Solution
-
Prepare the expression
Factor out \(\frac{1}{4}\):
\[
\left(\frac{1}{4} – 5x\right)^{\frac{1}{2}} = \left[\frac{1}{4}(1 – 20x)\right]^{\frac{1}{2}} = \left(\frac{1}{4}\right)^{\frac{1}{2}}(1 – 20x)^{\frac{1}{2}} = \frac{1}{2}(1 + (-20x))^{\frac{1}{2}}
\] -
Identify \(n\) and \(X\)
\(n = \frac{1}{2}\) and \(X = -20x\).
-
Apply the Binomial Formula
*Term in \(x^0: 1\)
*Term in \(x^1\):
\(nX = \left(\frac{1}{2}\right)(-20x) = -10x\)
*Term in \(x^2\):
\(\displaystyle \frac{n(n-1)}{2!}X^2 = \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}(-20x)^2 = -\frac{1}{8}(400x^2) = -50x^2\)
*Term in \(x^3\):
\(\displaystyle \frac{n(n-1)(n-2)}{3!}X^3 = \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}(-20x)^3 = \frac{1}{16}(-8000x^3) = -500x^3\) -
Multiply by \(\frac{1}{2}\):
\[
\frac{1}{2}(1 – 10x – 50x^2 – 500x^3 + \dots) = \frac{1}{2} – 5x – 25x^2 – 250x^3 + \dots
\]
Final Answer
\[\frac{1}{2} – 5x – 25x^2 – 250x^3 + …\]
Solution to Part b: Find an approximation for \(\sqrt{5}\) using \(x = \frac{1}{100}\).
Key Concepts Used
- Approximation: Substituting \(x\) to relate the function to the target value. .
- Algebraic Manipulation: Solving for \(\sqrt{5}\).
Step-by-Step Solution
-
Relate function to \(\sqrt{5}\)
Substitute \(x = \frac{1}{100}\) into the original function \(\left(\frac{1}{4} – 5x\right)^{\frac{1}{2}}\):
\[
\left(\frac{1}{4} – 5\left(\frac{1}{100}\right)\right)^{\frac{1}{2}} = \left(\frac{1}{4} – \frac{5}{100}\right)^{\frac{1}{2}} = \left(\frac{25}{100} – \frac{5}{100}\right)^{\frac{1}{2}} = \left(\frac{20}{100}\right)^{\frac{1}{2}} = \left(\frac{1}{5}\right)^{\frac{1}{2}} = \frac{1}{\sqrt{5}}
\] -
Calculate the approximation for \(\frac{1}{\sqrt{5}}\)
Substitute \(x = \frac{1}{100}\) into the expansion from (a), using the first 4 terms:
\[
\left(\frac{1}{4} – 5x\right)^{\frac{1}{2}} = \frac{1}{2} – 5x – 25x^2 – 250x^3 + \dots
\]
\[
\frac{1}{\sqrt{5}} \approx \frac{1}{2} – 5\left(\frac{1}{100}\right) – 25\left(\frac{1}{100}\right)^2 – 250\left(\frac{1}{100}\right)^3
\]
\[
\frac{1}{\sqrt{5}} = \frac{1}{2} – \frac{5}{100} – \frac{25}{10000} – \frac{250}{1000000}
\]
\[
\frac{1}{\sqrt{5}} = \frac{44725}{100000} = \frac{1789}{4000}
\] -
Find the approximation for \(\sqrt{5}\):
Take the reciprocal of the fraction:
\[
\sqrt{5} \approx \frac{4000}{1789}
\]
Final Answer
\[\sqrt{5} \approx \frac{4000}{1789} \quad (\text{where } a = 4000 \text{ and } b = 1789)\]