Practise Binomial expansion. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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Question 2Medium8 marks
A student is asked to find the binomial expansion of \((4 - x)^{-\frac{1}{2}}\) in ascending powers of \(x\), up to and including the term in \(x^2\), and to state the range of values of \(x\) for which the expansion is valid. The student writes:
\((4 - x)^{-\frac{1}{2}} = 4\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}} = 4\left(1 + \frac{x}{8} + \frac{3x^2}{128} + \ldots\right) = 4 + \frac{x}{2} + \frac{3x^2}{32} + \ldots\), valid for \(|x| \lt 1\).
(a) Identify the two errors in the student's work.[2]
(b) Write down the correct expansion, up to and including the term in \(x^2\).[2]
(c) By substituting \(x = 0.39\) into your expansion, find an approximation for \(\frac{10}{19}\). Give your answer to 6 decimal places.[3]
(d) Find the percentage error in this approximation.[1]
Show the answer and mark scheme
(a) Answer: The factor should be \(4^{-\frac{1}{2}} = \frac{1}{2}\), not 4; the expansion is valid for \(\left|\frac{x}{4}\right| \lt 1\), i.e. \(|x| \lt 4\), not \(|x| \lt 1\).
- B1 for the factor taken out should be \(4^{-\frac{1}{2}} = \frac{1}{2}\)
- B1 for the validity should be \(|x| \lt 4\) (from \(\left|\frac{x}{4}\right| \lt 1\))
Worked solution: \((4 - x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}\), so the factor is \(\frac{1}{2}\), not 4. The expansion of \((1 + u)^n\) is valid for \(|u| \lt 1\); here \(u = -\frac{x}{4}\), so it is valid for \(|x| \lt 4\).
(b) Answer: \(\frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256}\)
- M1 for \(\frac{1}{2}\left(1 + \frac{x}{8} + \frac{3x^2}{128}\right)\) (the student's bracket is correct)
- A1 for \(\frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256}\)
Worked solution: \(\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{x}{4}\right)^2 = 1 + \frac{x}{8} + \frac{3x^2}{128}\).
So \((4 - x)^{-\frac{1}{2}} \approx \frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256}\).
(c) Answer: 0.526157
- M1 for recognising \((4 - 0.39)^{-\frac{1}{2}} = 3.61^{-\frac{1}{2}} = \frac{1}{1.9} = \frac{10}{19}\)
- M1 for substituting \(x = 0.39\): \(0.5 + 0.024375 + 0.00178\ldots\)
- A1 for 0.526157
Worked solution: \(4 - 0.39 = 3.61 = 1.9^2\), so \((3.61)^{-\frac{1}{2}} = \dfrac{1}{1.9} = \dfrac{10}{19}\).
\(\frac{1}{2} + \frac{0.39}{16} + \frac{3(0.39)^2}{256} = 0.5 + 0.024375 + 0.001782\ldots = 0.526157\) (6 d.p.).
(d) Answer: 0.0301% (awrt 0.030%) %
- B1 for awrt 0.030% (from \(\frac{0.526316 - 0.526157}{0.526316} \times 100\))
Worked solution: \(\frac{10}{19} = 0.5263158\ldots\), so the percentage error is \(\dfrac{0.5263158 - 0.5261574}{0.5263158} \times 100 = 0.0301\%\).
Question 3Hard9 marks
(a) Find the binomial expansion of \(\sqrt{1 + 28x}\), in ascending powers of \(x\), up to and including the term in \(x^3\), simplifying each coefficient.[4]
(b) State the range of values of \(x\) for which the expansion is valid.[1]
(c) By substituting \(x = \frac{1}{100}\) into your expansion, find an approximation for \(\sqrt{2}\), giving your answer to 5 decimal places.[4]
Show the answer and mark scheme
(a) Answer: \(1 + 14x - 98x^2 + 1372x^3 + \ldots\)
- M1 for the correct structure of the \(x^2\) or \(x^3\) term, e.g. \(\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}(28x)^2\)
- B1 for \(1 + 14x\)
- A1 for \(-98x^2\)
- A1 for \(1372x^3\)
Worked solution: \((1 + 28x)^{\frac{1}{2}} = 1 + \frac{1}{2}(28x) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}(28x)^2 + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(28x)^3 + \ldots\)
\(= 1 + 14x - 98x^2 + 1372x^3 + \ldots\)
(b) Answer: \(|x| \lt \frac{1}{28}\)
- B1 for \(|x| \lt \frac{1}{28}\)
Worked solution: Valid for \(|28x| \lt 1\), i.e. \(|x| \lt \frac{1}{28}\).
(c) Answer: \(1.41447\)
- M1 for \(\sqrt{1 + \frac{28}{100}} = \frac{\sqrt{128}}{10}\)
- A1 for \(\frac{\sqrt{128}}{10} = \frac{8\sqrt{2}}{10}\) so \(\sqrt{2} = \frac{5}{4}\sqrt{1.28}\)
- M1 for substituting \(x = 0.01\) into their expansion and multiplying by \(\frac{5}{4}\)
- A1 for \(1.41447\)
Worked solution: \(\sqrt{1.28} = \sqrt{\frac{128}{100}} = \frac{\sqrt{64 \times 2}}{10} = \frac{8\sqrt{2}}{10}\), so \(\sqrt{2} = \frac{5}{4}\sqrt{1.28}\).
With \(x = 0.01\): \(1 + 14(0.01) - 98(0.01)^2 + 1372(0.01)^3 = 1.13157200\).
\(\sqrt{2} \approx \frac{5}{4} \times 1.13157200 = 1.4144650 = 1.41447\) (5 d.p.).