The shape of the lesson
By the end of the lesson, learners can
- Pascal's triangle and nCr: Multiply out \((a + b)^2\), \((a + b)^3\), \((a + b)^4\) and the coefficients build Pascal's triangle (Fig. 1).
- Expanding (a + b) n: Write the coefficients from \(\binom{n}{0}\) to \(\binom{n}{n}\); let the power of the first term fall from \(n\) to 0 while the power of the second rises from 0 to \(n\).
- Picking out one term: You rarely need the whole expansion.
- Products and unknowns: For a product such as \((1 + 2x)(1 - x)^6\), expand only as far as you need, then collect every pair of terms that multiplies to the power you want.
- Coefficients: \(\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}\) or Pascal's triangle.
- Powers: first term falls from \(n\), second rises from 0; they add to \(n\).
- Brackets: \((-3x)^2 = 9x^2\); keep the sign and the number inside the power.
- One term: simplify the power of \(x\) in the general term, then solve for \(r\).
- Products: list every pair that makes the power you want, and add.
One picture every idea returns to
\((a + b)^n = \displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r}\), where \(\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}\). The general term is \(\binom{n}{r}a^{n-r}b^{r}\): the powers of \(a\) and \(b\) always add up to \(n\).
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, Every bracket, expanded.. Multiplying out \((2 + x)^5\) by hand takes a page; the binomial theorem does it in one line. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: Pascal's triangle and nCr | Multiply out \((a + b)^2\), \((a + b)^3\), \((a + b)^4\) and the coefficients build Pascal's triangle (Fig. 1). Model the worked example Coefficients without a calculator on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Expanding (a + b) n | Write the coefficients from \(\binom{n}{0}\) to \(\binom{n}{n}\); let the power of the first term fall from \(n\) to 0 while the power of the second rises from 0 to \(n\). Model the worked example A full expansion on the board, one line at a time. Your turn: Find the first three terms, in ascending powers of \(x\), of \((3 + 2x)^5\). | Pairs |
| 17 to 23 min | Teach and check: Picking out one term | You rarely need the whole expansion. Model the worked example A single coefficient on the board, one line at a time. | Pairs |
| 23 to 30 min | Teach and check: Products and unknowns | For a product such as \((1 + 2x)(1 - x)^6\), expand only as far as you need, then collect every pair of terms that multiplies to the power you want. Model the worked example Coefficient in a product on the board, one line at a time. Your turn: Find the coefficient of \(x^2\) in the expansion of \((2 + x)^4(1 - 3x)\). | Round Table |
| 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, 20 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 Products and unknowns, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| \(\binom{5}{2}\) is not \(\dfrac{5!}{2!} = 60\). | Both factorials go underneath: \(\dfrac{5!}{2!\,3!} = 10\). |
| Bracket the whole second term before squaring: \((-3x)^2 = 9x^2\), not \(-3x^2\) and not \(-9x^2\). | Return to Expanding (a + b) n: ask a pair to state the correct version and show it on the lab before moving on. |
| The \(x^3\) term is the FOURTH term (\(r = 3\)), because counting starts at \(r = 0\). | Using \(\binom{6}{4}\) or \(3^4\) here is the most common slip. |
| Collect ALL the pairs. | In \((1 + 2x)(1 - x)^6\) the answer is not just 15: the \(2x\) multiplies the \(x\) term to make a second \(x^2\) term. |
The binomial expansion: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
(a) Find the value of \(\binom{8}{3}\).
(b) Show that \(\binom{n}{2} = \dfrac{n(n-1)}{2}\).
Expand \((2 - x)^4\) fully, simplifying each term.
Find the first three terms, in ascending powers of \(x\), in the expansion of \((3 + 2x)^5\).
Find the term independent of \(x\) in the expansion of \(\left(2x - \dfrac{1}{x^2}\right)^9\).
(a) Find the first three terms, in ascending powers of \(x\), of the expansion of \((2 - x)^6\).
(b) The coefficient of \(x^2\) in the expansion of \((1 + kx)(2 - x)^6\) is zero. Find the value of the constant \(k\).
A sum of $1000 is invested at 2% compound interest per year, so after 8 years it is worth \(1000(1 + 0.02)^8\) dollars.
(a) Find the first three terms, in ascending powers of \(x\), of \((1 + x)^8\).
(b) Use your answer to estimate the value of the investment after 8 years.
(c) Find the percentage error in your estimate, compared with the calculator value.
Answer key and marking notes
- (a) \(\dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)
- (b) \(\dfrac{n!}{2!\,(n-2)!} = \dfrac{n(n-1)(n-2)!}{2 \times (n-2)!} = \dfrac{n(n-1)}{2}\)
- Coefficients \(1, 4, 6, 4, 1\).
- \(2^4 + 4(2^3)(-x) + 6(2^2)(-x)^2 + 4(2)(-x)^3 + (-x)^4\)
- \(= 16 - 32x + 24x^2 - 8x^3 + x^4\)
- \(3^5 = 243\)
- \(\binom{5}{1}(3^4)(2x) = 5 \times 81 \times 2x = 810x\)
- \(\binom{5}{2}(3^3)(2x)^2 = 10 \times 27 \times 4x^2 = 1080x^2\)
- \(243 + 810x + 1080x^2\)
- General term: \(\binom{9}{r}(2x)^{9-r}\left(-\dfrac{1}{x^2}\right)^{r} = \binom{9}{r}2^{9-r}(-1)^{r}x^{9-3r}\)
- \(9 - 3r = 0\), so \(r = 3\).
- \(\binom{9}{3} \times 2^6 \times (-1)^3 = 84 \times 64 \times (-1) = -5376\)
- (a) \(64 + 6(32)(-x) + 15(16)x^2 = 64 - 192x + 240x^2\)
- (b) \(x^2\) terms: \(1 \times 240x^2\) and \(kx \times (-192x)\).
- \(240 - 192k = 0\), so \(k = \dfrac{240}{192} = \dfrac{5}{4}\)
- (a) \(1 + 8x + 28x^2\)
- (b) \(x = 0.02\): \(1 + 0.16 + 28 \times 0.0004 = 1.1712\), so about $1171.20.
- (c) Calculator: \(1000 \times 1.02^8 = 1171.66\).
- Error \(= \dfrac{1171.66 - 1171.20}{1171.66} \times 100 = 0.0392\%\) (3 s.f.); the neglected terms are tiny because \(x\) is small.
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.
Find the value of \(\dbinom{8}{3}\).
- A56
- B336
- C24
- D70
Find the first three terms, in ascending powers of \(x\), of the expansion of \(\left(1 + \dfrac{x}{2}\right)^8\).
- A\(1 + 4x + 7x^2\)
- B\(1 + 4x + 14x^2\)
- C\(1 + 4x + 28x^2\)
- D\(1 + 8x + 28x^2\)
Find the coefficient of \(x^4\) in the expansion of \((1 - 2x)^6\).
- A240
- B\(-240\)
- C30
- D15
Find the term in \(x^3\) in the expansion of \((2 + x)^5\).
- A\(40x^3\)
- B\(80x^3\)
- C\(10x^3\)
- D\(20x^3\)
Find the term independent of \(x\) in the expansion of \(\left(2x - \dfrac{1}{x^2}\right)^9\).
- A\(-5376\)
- B\(5376\)
- C\(-672\)
- D\(-84\)
Find the coefficient of \(x^2\) in the expansion of \((2 - x)(1 + 3x)^5\).
- A165
- B195
- C180
- D75
In the expansion of \((1 + ax)^7\), where \(a > 0\), the coefficient of \(x^3\) is 10 times the coefficient of \(x\). Find \(a\).
- A\(\sqrt{2}\)
- B\(2\)
- C\(\sqrt{\dfrac{2}{7}}\)
- D\(2\sqrt{2}\)
The first three terms of the expansion of \((1 + kx)^n\), in ascending powers of \(x\), are \(1 + 12x + 60x^2\). Find \(n\) and \(k\).
- A\(n = 6,\ k = 2\)
- B\(n = 12,\ k = 1\)
- C\(n = 4,\ k = 3\)
- D\(n = 3,\ k = 4\)
Quiz key
\(\dbinom{8}{3} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = \dfrac{336}{6} = 56\). The tempting 336 is \(8 \times 7 \times 6\) without dividing by \(3!\) (it counts ordered choices). 24 is \(8 \times 3\), and 70 is \(\dbinom{8}{4}\), the middle of row 8 of Pascal's triangle.
\(1 + 8\left(\dfrac{x}{2}\right) + 28\left(\dfrac{x}{2}\right)^2 = 1 + 4x + 28 \times \dfrac{x^2}{4} = 1 + 4x + 7x^2\). The tempting \(14x^2\) squares the \(x\) but not the \(\tfrac{1}{2}\). \(28x^2\) forgets the \(\tfrac{1}{2}\) altogether in the third term, and \(1 + 8x + 28x^2\) is the expansion of \((1 + x)^8\).
\(\dbinom{6}{4}(-2)^4 = 15 \times 16 = 240\). The tempting \(-240\) keeps the minus sign, but an even power of a negative number is positive. 30 uses \(-2\) without the power (\(15 \times 2\)), and 15 ignores the 2 completely.
The term is \(\dbinom{5}{3} \times 2^{2} \times x^3 = 10 \times 4x^3 = 40x^3\). The tempting \(80x^3\) gives the 2 the same power as \(x\) (\(2^3\)), but the powers must add to 5. \(10x^3\) forgets the powers of 2 altogether, and \(20x^3\) uses \(2^1\).
The power of \(x\) is \((9 - r) - 2r = 9 - 3r\), which is 0 when \(r = 3\). The term is \(\dbinom{9}{3}(2x)^6\left(-\dfrac{1}{x^2}\right)^3 = 84 \times 64 \times (-1) = -5376\). The tempting \(5376\) loses the sign: \((-1)^3 = -1\). \(-672\) uses \(2^3\) instead of \(2^6\), and \(-84\) forgets the powers of 2.
\((1 + 3x)^5 = 1 + 15x + 90x^2 + \cdots\). The \(x^2\) terms in the product are \(2 \times 90x^2\) and \(-x \times 15x\), so the coefficient is \(180 - 15 = 165\). The tempting 195 adds the 15 instead of subtracting it (the \(-x\) carries a minus sign). 180 forgets the \(-x \times 15x\) product, and 75 uses 1 instead of 2 in front of the 90.
\(\dbinom{7}{3}a^3 = 10 \times 7a\) gives \(35a^3 = 70a\). Since \(a \ne 0\), \(a^2 = 2\), and \(a > 0\) so \(a = \sqrt{2}\). The tempting 2 stops at \(a^2 = 2\) without taking the square root. \(\sqrt{\dfrac{2}{7}}\) takes the coefficient of \(x\) as \(a\) instead of \(7a\), and \(2\sqrt{2}\) is \(a^3\).
\(\dfrac{n(n-1)}{2}k^2 = \dfrac{(nk)(n-1)k}{2} = 6(n - 1)k = 60\), so \((n - 1)k = 10\), that is \(nk - k = 10\), giving \(k = 2\) and \(n = 6\). Check: \(\dbinom{6}{2} \times 4 = 60\). The other pairs all have \(nk = 12\) but give the wrong \(x^2\) coefficient: \(n = 12, k = 1\) gives \(\dbinom{12}{2} = 66\), \(n = 4, k = 3\) gives \(6 \times 9 = 54\), and \(n = 3, k = 4\) gives \(3 \times 16 = 48\). Matching only the \(x\) term is the tempting mistake.
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.