Edexcel IAL June 2021 P4 (WMA14/01) Q1 Binomial Expansion & Approximation
Given that \( k \) is a constant and the binomial expansion of
\[\sqrt{1 + kx} \quad |kx| < 1\]
in ascending powers of \( x \) up to the term in \( x^3 \) is
\[1 + \frac{1}{8}x + Ax^2 + Bx^3\]
(a) (i) find the value of \( k \),
(ii) find the value of the constant \( A \) and the constant \( B \).
(5)
(b) Use the expansion to find an approximate value to \(\sqrt{1.15}\)
(5)
Show your working and give your answer to 6 decimal places.
(2)
Solution to Part a (i): Find the value of \( k \).
Key Concepts Used
- Series Comparison: Equating the coefficient of \( x \) in the expansion to the given value, \(\frac{1}{8}\).
Step-by-Step Solution
-
Expand the \( x \) term:
\[\sqrt{1 + kx} = (1 + kx)^{1/2} \approx 1 + \left(\frac{1}{2}\right)(kx) + \cdots\]
\[ \text{Coeff. of } x = \frac{k}{2} \] -
Equate and solve
Set the derived coefficient equal to the given coefficient,
\[ \frac{k}{2} = \frac{1}{8} \Rightarrow k = \frac{2}{8} = \frac{1}{4} \]
Final Answer
\[ k = \frac{1}{4} \]
Solution to Part a (ii): Find the value of \( A \) and \( B \).
Key Concepts Used
- Coefficient Calculation: Substitute the value of \( k \) into the coefficient formulas for \( x^2 \) and \( x^3 \).
Step-by-Step Solution (Part a(ii))
-
Calculate \(A\) (coefficient of \(x^{2}\)):
\[A=\frac{n(n-1)}{2!}k^{2}=\frac{(\frac{1}{2})(-\frac{1}{2})}{2}k^{2}=-\frac{1}{8 }k^{2}\]
Substitute \(k=\frac{1}{4}\):
\[A=-\frac{1}{8}\Big{(}\frac{1}{4}\Big{)}^{2}=-\frac{1}{8}\Big{(}\frac{1}{16 }\Big{)}=-\frac{1}{128}\] -
Calculate \(B\) (coefficient of \(x^{3}\)):
\[B=\frac{n(n-1)(n-2)}{3!}k^{3}=\frac{(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})}{6 }k^{3}=\frac{1}{16}k^{3\]
Substitute \(k=\frac{1}{4}\):
\[B=\frac{1}{16}\Big{(}\frac{1}{4}\Big{)}^{3}=\frac{1}{16}\Big{(}\frac{1}{64}\Big{ )}=\frac{1}{1024}\]
Final Answer
\[A=-\frac{1}{128} \quad B=\frac{1}{1024}\]
Solution to Part b: Use the expansion to find an approximate value to \(\sqrt{1.15}\).
Key Concepts Used
- Approximation: Find the value of \(x\) that relates the expression to \(\sqrt{1.15}\).
- Substitution: Evaluate the expansion at this \(x\) value.
Step-by-Step Solution
-
Find the required \(x\)
Set the argument of the square root equal to 1.15:
\[1+kx=1+\frac{1}{4}x=1.15\]
\[\frac{1}{4}x=0.15 \Rightarrow x=4(0.15)=0.6\] -
Substitute \(x=0.6\) into the expansion
Use the expansion up to the \(x^{3}\) term:
\[\sqrt{1.15}\approx 1+\frac{1}{8}(0.6)+A(0.6)^{2}+B(0.6)^{3}\]
\[\approx 1+\frac{1}{8}(0.6)-\frac{1}{128}(0.6)^{2}+\frac{1}{1024}(0.6)^{3\]
\[\approx 1+0.075-0.0028125+0.0002109375\]
\[\approx 1.0723984375\] -
Round to 6 decimal places
1.072398
Final Answer
\[\sqrt{1.15} \approx 1.072398\]