Binomial Expansion

Edexcel IAL October 2020 P4 (WMA14/01) Q2 Binomial Expansion

(a) Use the binomial expansion to expand

\[(4 – 5x)^{-\frac{1}{2}} \quad |x| < \frac{4}{5}\]

in ascending powers of $x$, up to and including the term in $x^2$, giving each coefficient as a fully simplified fraction.

(4)

\[f(x) = \frac{2 + kx}{\sqrt{4 – 5x}} \quad \text{where } k \text{ is a constant and } |x| < \frac{4}{5}\]

Given that the series expansion of $f(x)$, in ascending powers of $x$, is

\[1 + \frac{3}{10}x + mx^2 + \dots \quad \text{where } m \text{ is a constant}\]

(b) find the value of $k$,

(2)

Solution to Part a: Expand $(4 – 5x)^{-\frac{1}{2}}$ up to $x^2$.

Key Concepts Used:

Step-by-Step Solution:

$Prepare the expression$

Factor out 4 and rewrite the root as a power
\[(4 – 5x)^{-\frac{1}{2}} = [4(1 – \frac{5}{4}x)]^{-\frac{1}{2}} = 4^{-\frac{1}{2}}(1 – \frac{5}{4}x)^{-\frac{1}{2}} = \frac{1}{2}(1 + (-\frac{5}{4}x))^{-\frac{1}{2}}\]
$Identify $n$ and $X$:$

$n = -\frac{1}{2}$ and $X = -\frac{5}{4}x$.
$Apply the Binomial Formula$

\[(1 – \frac{5}{4}x)^{-\frac{1}{2}} \approx 1 + (-\frac{1}{2})(-\frac{5}{4}x) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(-\frac{5}{4}x)^2\]
\[\approx 1 + \frac{5}{8}x + \frac{3}{4}(\frac{25}{16}x^2) = 1 + \frac{5}{8}x + \frac{75}{128}x^2 + \dots\]
$Multiply by $\frac{1}{2}$$

\[\frac{1}{2} \times \left(1 + \frac{5}{8}x + \frac{75}{128}x^2 + \dots\right) = \frac{1}{2} + \frac{5}{16}x + \frac{75}{256}x^2 + \dots\]

Final Answer:

\[(4 – 5x)^{-\frac{1}{2}} \approx \frac{1}{2} + \frac{5}{16}x + \frac{75}{256}x^2 + \dots\]

Solution to Part b: Find the value of $ k $.

Key Concepts Used:

Series Comparison: Equating the coefficient of $ x $ in the full expansion of $ f(x) $ with the given series coefficient $ \frac{3}{10} $.

Step-by-Step Solution:

$Set up $ f(x) $$

\[f(x) = (2 + kx)\left(\frac{1}{2} + \frac{5}{16}x + \dots\right)\]
$Find the constant term$

The constant term is $ 2 \times \frac{1}{2} = 1 $.
This matches the given expansion $ (1 + \frac{3}{10}x + mx^2 + \dots) $ given in the question.
$Find the coefficient of $ x $$

This is the sum of the $ x $ terms generated by multiplying the constants and the $ x $ terms in the brackets:
\[\text{Coeff. of } x = (2) \times \left(\frac{5}{16}\right) + (kx) \times \left(\frac{1}{2}\right) = \frac{10}{16} + \frac{k}{2}\]
$Equate coefficients and solve for $ k $$

Set this equal to the given coefficient of $ x $:
\[\frac{10}{16} + \frac{k}{2} = \frac{3}{10} \implies \frac{5}{8} + \frac{k}{2} = \frac{3}{10}\]
\[\frac{k}{2} = \frac{3}{10} – \frac{5}{8} = \frac{12}{40} – \frac{25}{40} = -\frac{13}{40}\]
\[k = 2 \times \left(-\frac{13}{40}\right) = -\frac{13}{20}\]

Final Answer:

\[k = -\frac{13}{20}\]

Solution to Part c: Find the value of $ m $.

Key Concepts Used:

Series Comparison: Equating the coefficient of $ x^2 $ in the full expansion of $ f(x) $ with the given constant $ m $.

Step-by-Step Solution:

$Find the coefficient of $ x^2 $$

This is the sum of the $ x^2 $ terms generated by multiplying the constant term (2) by the $ x^2 $ term, and the $ kx $ term by the $ \frac{5}{16} x $ term:
\[m = \text{Coeff. of } x^2 = (2) \times \left( \frac{75}{256} \right) + (k) \times \left( \frac{5}{16} \right)\]
$Substitute $ k = -\frac{13}{20} $$

\[m = \frac{150}{256} + \left( -\frac{13}{20} \right) \times \left( \frac{5}{16} \right)\]
$Simplify$

\[m = \frac{75}{128} – \frac{65}{320} = \frac{75}{128} – \frac{13}{64}\]
\[m = \frac{75}{128} – \frac{26}{128} = \frac{49}{128}\]

Final Answer:

\[m = \frac{49}{128}\]

Related Course