The shape of the lesson
By the end of the lesson, learners can
- The series for any rational \(n\): For a positive integer \(n\) the expansion stops after \(n + 1\) terms.
- Adapting to \((a + bx)^n\): take out \(a^n\): The standard series needs a 1 in front.
- With partial fractions: A rational function is expanded by splitting it into partial fractions, expanding each piece, and adding.
- Standard form: \((a + bx)^n = a^n\left(1 + \dfrac{b}{a}x\right)^n\).
- Coefficients: \(1,\ n,\ \dfrac{n(n - 1)}{2!},\ \dfrac{n(n - 1)(n - 2)}{3!}\).
- Brackets: raise the whole term, sign and number, to each power.
- Validity: \(\left|\dfrac{bx}{a}\right| < 1\); for sums take the narrowest range.
- Products: collect every pair of terms that multiply to the power you need.
One picture every idea returns to
\((1 + x)^n = 1 + nx + \dfrac{n(n - 1)}{2!}x^2 + \dfrac{n(n - 1)(n - 2)}{3!}x^3 + \dots\), valid for \(|x| < 1\) when \(n\) is not a positive integer. For \((a + bx)^n\), first take out \(a^n\): \(a^n\left(1 + \dfrac{b}{a}x\right)^n\), valid for \(\left|\dfrac{bx}{a}\right| < 1\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Infinite, but useful.. Paper 3 only. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The series for any rational \(n\) | For a positive integer \(n\) the expansion stops after \(n + 1\) terms. Model the worked example A negative index on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Adapting to \((a + bx)^n\): take out \(a^n\) | The standard series needs a 1 in front. Model the worked example Taking out a constant on the board, one line at a time. Your turn: Expand \(\dfrac{1}{\sqrt{4 + x}}\) up to the term in \(x^2\), and state the values of \(x\) for which it is valid. | Pairs |
| 21 to 30 min | Teach and check: With partial fractions | A rational function is expanded by splitting it into partial fractions, expanding each piece, and adding. Model the worked example Split, expand, add on the board, one line at a time. | Jigsaw |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 25 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in With partial fractions, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \((-4x)^2 = 16x^2\), not \(-4x^2\) or \(4x^2\). | Write the bracket in every term, then square or cube both the number and the sign. |
| \((2 - x)^{-3}\) is NOT \(2\left(1 - \dfrac{x}{2}\right)^{-3}\). | The factor taken out is \(2^{-3} = \dfrac{1}{8}\): it carries the same power as the bracket. |
| Write \(\dfrac{2}{2x - 1}\) as \(-2(1 - 2x)^{-1}\) before expanding: the standard series needs the constant term in the bracket to be \(+1\). | Expanding \((2x - 1)^{-1}\) 'as if' it were \((1 + 2x)^{-1}\) gives the wrong signs. |
The binomial series for any rational n: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
Expand \((1 - 2x)^{-1/2}\) in ascending powers of \(x\), up to and including the term in \(x^3\).
Expand \(\sqrt[3]{8 + 3x}\) up to the term in \(x^2\), and state the set of values of \(x\) for which the expansion is valid.
Find the coefficient of \(x^3\) in the expansion of \((1 + 3x)^{1/3}\).
(a) Expand \((1 + x)^{1/2}\) up to the term in \(x^2\).
(b) By substituting \(x = \dfrac{1}{16}\), find an approximation to \(\sqrt{17}\), giving your answer to 4 decimal places.
In the expansion of \((1 + ax)^{n}\), the coefficient of \(x\) is \(-6\) and the coefficient of \(x^2\) is \(27\). Find \(n\) and \(a\).
Let \(f(x) = \dfrac{5x^2 + 3x + 7}{(x + 2)(x^2 + 3)}\).
(a) Given that \(f(x) \equiv \dfrac{3}{x + 2} + \dfrac{2x - 1}{x^2 + 3}\), find the expansion of \(f(x)\) in ascending powers of \(x\), up to and including the term in \(x^2\).
(b) State the set of values of \(x\) for which the expansion is valid.
Answer key and marking notes
- \(x\): \(\left(-\dfrac{1}{2}\right)(-2x) = x\).
- \(x^2\): \(\dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}(4x^2) = \dfrac{3}{8} \cdot 4x^2 = \dfrac{3}{2}x^2\).
- \(x^3\): \(\dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}(-8x^3) = -\dfrac{5}{16} \cdot (-8x^3) = \dfrac{5}{2}x^3\).
- \(1 + x + \dfrac{3}{2}x^2 + \dfrac{5}{2}x^3\), valid for \(|x| < \dfrac{1}{2}\).
- \((8 + 3x)^{1/3} = 8^{1/3}\left(1 + \dfrac{3x}{8}\right)^{1/3} = 2\left(1 + \dfrac{3x}{8}\right)^{1/3}\).
- \(= 2\left(1 + \dfrac{1}{3} \cdot \dfrac{3x}{8} + \dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2} \cdot \dfrac{9x^2}{64}\right) = 2\left(1 + \dfrac{x}{8} - \dfrac{x^2}{64}\right)\).
- \(= 2 + \dfrac{x}{4} - \dfrac{x^2}{32}\).
- Valid for \(\left|\dfrac{3x}{8}\right| < 1\), i.e. \(|x| < \dfrac{8}{3}\).
- Coefficient: \(\dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{6} \times 3^3 = \dfrac{10}{27 \times 6} \times 27\).
- \(= \dfrac{10}{6} = \dfrac{5}{3}\).
- (a) \(1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2\), valid for \(|x| < 1\).
- (b) \(\sqrt{1 + \dfrac{1}{16}} = \sqrt{\dfrac{17}{16}} = \dfrac{\sqrt{17}}{4}\).
- Series: \(1 + \dfrac{1}{32} - \dfrac{1}{2048} = 1.030\,761\dots\)
- \(\sqrt{17} \approx 4 \times 1.030\,761 = 4.1230\) (4 d.p.). The true value is \(4.1231\dots\)
- \(na = -6\) and \(\dfrac{n(n - 1)}{2}a^2 = 27\).
- Write the second as \(\dfrac{(na)^2}{2} \cdot \dfrac{n - 1}{n} = 27\): \(18 \cdot \dfrac{n - 1}{n} = 27\).
- \(18n - 18 = 27n\), so \(n = -2\).
- \(a = \dfrac{-6}{n} = 3\). (Check: \((1 + 3x)^{-2} = 1 - 6x + 27x^2 - \dots\))
- (a) \(\dfrac{3}{x + 2} = \dfrac{3}{2}\left(1 + \dfrac{x}{2}\right)^{-1} = \dfrac{3}{2}\left(1 - \dfrac{x}{2} + \dfrac{x^2}{4}\right) = \dfrac{3}{2} - \dfrac{3}{4}x + \dfrac{3}{8}x^2\).
- \(\dfrac{1}{x^2 + 3} = \dfrac{1}{3}\left(1 + \dfrac{x^2}{3}\right)^{-1} = \dfrac{1}{3} - \dfrac{x^2}{9} + \dots\)
- \((2x - 1)\left(\dfrac{1}{3} - \dfrac{x^2}{9}\right) = -\dfrac{1}{3} + \dfrac{2}{3}x + \dfrac{1}{9}x^2 + \dots\) (terms up to \(x^2\)).
- Add: \(\left(\dfrac{3}{2} - \dfrac{1}{3}\right) + \left(-\dfrac{3}{4} + \dfrac{2}{3}\right)x + \left(\dfrac{3}{8} + \dfrac{1}{9}\right)x^2 = \dfrac{7}{6} - \dfrac{1}{12}x + \dfrac{35}{72}x^2\).
- (b) First piece: \(|x| < 2\). Second: \(\dfrac{x^2}{3} < 1\), i.e. \(|x| < \sqrt{3}\). Both: \(|x| < \sqrt{3}\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
(Paper 3 only) Find the first three terms in the expansion of \((1 + x)^{-4}\) in ascending powers of \(x\).
- A\(1 - 4x + 10x^2\)
- B\(1 - 4x + 20x^2\)
- C\(1 + 4x + 10x^2\)
- D\(1 - 4x + 6x^2\)
(Paper 3 only) For which values of \(x\) is the expansion of \((2 + 3x)^{-\frac{1}{2}}\) in ascending powers of \(x\) valid?
- A\(|x| < \dfrac{2}{3}\)
- B\(|x| < \dfrac{3}{2}\)
- C\(|x| < 1\)
- D\(|x| < \dfrac{1}{3}\)
(Paper 3 only) Expand \(\sqrt{1 + 4x}\) in ascending powers of \(x\), up to and including the term in \(x^2\).
- A\(1 + 2x - 2x^2\)
- B\(1 + 2x - \dfrac{1}{2}x^2\)
- C\(1 + 2x + 2x^2\)
- D\(1 + 2x - 4x^2\)
(Paper 3 only) Find the coefficient of \(x^3\) in the expansion of \((1 - x)^{-2}\).
- A\(4\)
- B\(-4\)
- C\(24\)
- D\(3\)
(Paper 3 only) Expand \((2 - x)^{-2}\) in ascending powers of \(x\), up to and including the term in \(x^2\).
- A\(\dfrac{1}{4} + \dfrac{1}{4}x + \dfrac{3}{16}x^2\)
- B\(\dfrac{1}{4} + x + \dfrac{3}{4}x^2\)
- C\(4 + 4x + 3x^2\)
- D\(\dfrac{1}{4} - \dfrac{1}{4}x + \dfrac{3}{16}x^2\)
(Paper 3 only) The expansion of \(\sqrt{1 - 2x}\) up to the term in \(x^2\) is \(1 - x - \dfrac{1}{2}x^2\). Use it with \(x = 0.02\) to find an approximation for \(\sqrt{96}\).
- A\(9.798\)
- B\(9.8\)
- C\(9.802\)
- D\(0.9798\)
(Paper 3 only) Given that \(f(x) \equiv \dfrac{2}{1 - x} + \dfrac{3}{2 + x}\), expand \(f(x)\) in ascending powers of \(x\), up to and including the term in \(x^2\).
- A\(\dfrac{7}{2} + \dfrac{5}{4}x + \dfrac{19}{8}x^2\)
- B\(5 + \dfrac{1}{2}x + \dfrac{11}{4}x^2\)
- C\(\dfrac{7}{2} + \dfrac{1}{2}x + \dfrac{7}{2}x^2\)
- D\(\dfrac{7}{2} - \dfrac{11}{4}x + \dfrac{19}{8}x^2\)
(Paper 3 only) In the expansion of \((1 + ax)^{-3}\), the coefficient of \(x^2\) is 24. Find the possible values of \(a\).
- A\(a = \pm 2\)
- B\(a = 2\) only
- C\(a = \pm\sqrt{2}\)
- D\(a = 4\)
Quiz key
\(nx = -4x\) and \(\dfrac{(-4)(-5)}{2}x^2 = 10x^2\), so \(1 - 4x + 10x^2\). The tempting \(20x^2\) forgets to divide by \(2!\). \(1 + 4x\) loses the sign of \(n\), and \(6x^2\) uses \((-4)(-3)\), i.e. \(n + 1\) instead of \(n - 1\).
\((2 + 3x)^{-\frac{1}{2}} = 2^{-\frac{1}{2}}\left(1 + \dfrac{3x}{2}\right)^{-\frac{1}{2}}\), valid for \(\left|\dfrac{3x}{2}\right| < 1\), i.e. \(|x| < \dfrac{2}{3}\). The tempting \(|x| < \dfrac{3}{2}\) inverts the fraction. \(|x| < 1\) applies the rule for \((1 + x)^n\) without adapting it, and \(|x| < \dfrac{1}{3}\) ignores the 2 that was taken out.
\(\dfrac{1}{2}(4x) = 2x\); \(\dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}(4x)^2 = -\dfrac{1}{8}(16x^2) = -2x^2\). So \(1 + 2x - 2x^2\). The tempting \(-\dfrac{1}{2}x^2\) uses \(4x^2\) instead of \((4x)^2 = 16x^2\). \(+2x^2\) loses the negative sign of \(n - 1 = -\dfrac{1}{2}\), and \(-4x^2\) forgets to divide by \(2!\).
\(\dfrac{(-2)(-3)(-4)}{6}(-x)^3 = (-4)(-x^3) = 4x^3\). (In fact \((1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \dots\).) The tempting \(-4\) uses \(x^3\) instead of \((-x)^3\). 24 forgets to divide by \(3! = 6\), and 3 is the coefficient of \(x^2\).
\(\left(1 - \dfrac{x}{2}\right)^{-2} = 1 + (-2)\left(-\dfrac{x}{2}\right) + \dfrac{(-2)(-3)}{2}\left(-\dfrac{x}{2}\right)^2 = 1 + x + \dfrac{3}{4}x^2\). Times \(\dfrac{1}{4}\): \(\dfrac{1}{4} + \dfrac{1}{4}x + \dfrac{3}{16}x^2\). The tempting \(\dfrac{1}{4} + x + \dfrac{3}{4}x^2\) multiplies only the first term by \(\dfrac{1}{4}\). \(4 + 4x + 3x^2\) uses \(2^{2}\) instead of \(2^{-2}\), and \(-\dfrac{1}{4}x\) drops one of the two minus signs in \((-2)\left(-\dfrac{x}{2}\right)\).
\(\sqrt{0.96} \approx 1 - 0.02 - \dfrac{1}{2}(0.0004) = 0.9798\). Since \(96 = 100 \times 0.96\), \(\sqrt{96} = 10\sqrt{0.96} \approx 9.798\) (calculator: \(9.79796\ldots\)). The tempting 9.8 uses only two terms. 9.802 adds the \(x^2\) term instead of subtracting it, and 0.9798 forgets the factor \(\sqrt{100} = 10\).
\(2(1 + x + x^2) = 2 + 2x + 2x^2\). \(\dfrac{3}{2}\left(1 - \dfrac{x}{2} + \dfrac{x^2}{4}\right) = \dfrac{3}{2} - \dfrac{3}{4}x + \dfrac{3}{8}x^2\). Sum: \(\dfrac{7}{2} + \dfrac{5}{4}x + \dfrac{19}{8}x^2\). The tempting \(5 + \dfrac{1}{2}x + \dfrac{11}{4}x^2\) expands \(3\left(1 + \dfrac{x}{2}\right)^{-1}\), forgetting the factor \(\dfrac{1}{2}\) from \(2^{-1}\). \(\dfrac{7}{2} + \dfrac{1}{2}x + \dfrac{7}{2}x^2\) expands \((2 + x)^{-1}\) as \(\dfrac{1}{2}(1 - x + x^2)\), not halving \(x\) inside, and \(-\dfrac{11}{4}x\) expands \((1 - x)^{-1}\) as \(1 - x + x^2\).
\(\dfrac{(-3)(-4)}{2}a^2x^2 = 6a^2x^2\), so \(6a^2 = 24\), \(a^2 = 4\), \(a = \pm 2\). Both work, because \(a^2\) is the same for each. The tempting "\(a = 2\) only" forgets the negative root. \(\pm\sqrt{2}\) forgets to divide by \(2!\) (\(12a^2 = 24\)), and \(a = 4\) uses \(ax^2\) instead of \((ax)^2\), giving \(6a = 24\).
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.