Series

• One line answers every “find the coefficient” question: the term containing \(b^r\) is \(\binom{n}{r}a^{n-r}b^r\). One condition guards every sum to infinity: \(|r| < 1\), checked before the formula is written.
• Three lines that mean the chapter has stuck: the term containing \(b^r\) is \(\binom{n}{r}a^{n-r}b^r\); two facts make two equations in \(a\) and \(d\), or \(a\) and \(r\); and \(S_\infty = \dfrac{a}{1-r}\) only when \(|r| < 1\).

• 1.6.1 I can use — use the expansion of (a + b)ⁿ, where n is a positive integer
• 1.6.2 I can recognise — recognise arithmetic and geometric progressions
• 1.6.3 I can use — use the formulae for the nth term and for the sum of the first n terms to solve problems involving arithmetic or geometric progressions
• 1.6.4 I can use — use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression

Accuracy on this chapter. Coefficients and the sums of progressions are exact: write \(-160\), \(765\), \(\tfrac{32}{3}\), \(\tfrac{3}{11}\), not decimals. A common ratio found from a cube root is exact (\(\tfrac23\)), and an \(S_\infty\) stays a fraction unless the question says otherwise. A least \(n\) is an integer taken above an unrounded critical value (17.13, so \(n = 18\)) and confirmed by \(S_n\) either side. Where a question gives money or lengths, round the final answer only, to 3 significant figures.

• “The coefficient of \(x^3\) in \((1 + 2x)^7\) is \(\binom{7}{3} = 35\), and it is the third term.” Correct The term containing \(x^r\) is the \((r+1)\)th term, because the count starts at \(r = 0\): the \(x^3\) term is the fourth. And its coefficient includes the power of the number in front of \(x\): \(\binom{7}{3}(2x)^3 = 35 \times 8x^3\), coefficient 280. (Lesson card A.)
• “\(a = 5\), \(r = 2\), so \(S_\infty = \dfrac{5}{1 - 2} = -5\).” Correct \(S_\infty\) exists only when \(|r| < 1\). With \(r = 2\) the terms 5, 10, 20, … grow without limit and there is no sum to infinity; the formula must not be used outside its condition. Check \(-1 < r < 1\) before writing \(\dfrac{a}{1-r}\). (Lesson card D.)
• “\((2 - 3x)^4\) has an \(x^3\) term of \(4 \times 2 \times 27x^3 = 216x^3\).” Correct Here \(b = -3x\), sign included, and the whole of \(b\) is cubed in brackets: \((-3x)^3 = -27x^3\), so the term is \(-216x^3\). Odd powers of a negative \(b\) are negative. (Lesson card A.)
• “2, \(x\), 18 are in geometric progression, so \(x = 6\).” Correct \(b^2 = ac\) gives \(x^2 = 36\), so \(x = \pm 6\): 2, \(-6\), 18 is geometric too, with \(r = -3\). Keep both unless the question rules one out. (Lesson card B.)
• “\(S_n > 500\) gives \(n = 17.13\), so \(n = 17\).” Correct \(n\) counts terms, so it is an integer, and the sum first exceeds 500 at the next integer above the critical value: \(n = 18\). Confirm it: \(S_{17} = 493 < 500 < 549 = S_{18}\). (Lesson card C.)
• “\(1 + (x - 2) + (x - 2)^2 + \cdots\) converges when \(x - 2 < 1\), that is \(x < 3\).” Correct The condition is two-sided, \(-1 < x - 2 < 1\), so the answer is \(1 < x < 3\). At \(x = 0\), say, \(r = -2\) and the progression does not converge. (Lesson card D.)
• “\((2 - 3x)^4 = 16 - 96x + 216x^2 + 216x^3 + 81x^4\).” Repair Keep the sign inside the bracket: \((-3x)^3 = -27x^3\), so the \(x^3\) term is \(4 \times 2 \times (-27x^3) = -216x^3\). The \(x = 1\) check catches it: the wrong version sums to 433, not 1.
• “The third term of \((1 + 2x)^7\) is \(\binom{7}{3}(2x)^3\).” Repair The term containing \(x^3\) is the fourth term: \(\binom{7}{r}\) goes with \(x^r\), and the count starts at \(r = 0\). The third term is \(\binom{7}{2}(2x)^2 = 84x^2\).
• “The coefficient of \(x^3\) in \((1 + 2x)^7\) is 35.” Repair The 2 is raised to the power too: \(35 \times 2^3 = 280\).
• “\(\binom{7}{3} = \dfrac{7!}{3!} = 840\).” Repair \(\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}\): both factorials go in the denominator. \(\dfrac{5040}{6 \times 24} = 35\).
• “\(10a^2 = 90\), so \(a = 3\).” Repair \(a^2 = 9\) gives \(a = \pm 3\). Only a condition in the question, such as \(a > 0\), allows \(-3\) to be discarded.
• “The coefficient of \(x^2\) in \((1 + 2x)(1 - x)^6\) is 15.” Repair Two pairings make \(x^2\): \(1 \times 15x^2\) and \(2x \times (-6x)\). The coefficient is \(15 - 12 = 3\).
• “1, 4, 9, 16 is an AP because it goes up each time.” Repair The differences are 3, 5, 7, not constant, so it is not an AP; the ratios 4, 2.25, 1.78 are not constant either, so it is not a GP.
• “2, \(x\), 18 are in GP, so \(x = 6\).” Repair \(x^2 = 36\) gives \(x = \pm 6\); 2, \(-6\), 18 is also geometric, with \(r = -3\).
• “\(u_{10} = a + 10d\).” Repair \(u_n = a + (n-1)d\), so \(u_{10} = a + 9d\). Ten terms have nine gaps.
• “\(S_{10} = 185\), so \(10(2a + 9d) = 185\).” Repair The \(\tfrac12\) is part of the formula: \(\tfrac12(10)(2a + 9d) = 5(2a + 9d) = 185\).
• “\(S_n > 500\) gives \(n = 17.13\), so \(n = 17\).” Repair \(n\) is an integer and \(S_n\) is increasing here, so take the next integer above the critical value, \(n = 18\), and check \(S_{17} = 493 < 500 < 549 = S_{18}\).
• “\(S_\infty = \dfrac{a}{1 - r} = \dfrac{5}{1 - 2} = -5\).” Repair With \(r = 2\) the progression does not converge. \(S_\infty\) exists only for \(|r| < 1\), and the formula must not be used outside that condition.
• “The GP converges when \(r < 1\).” Repair The condition is \(-1 < r < 1\). \(r = -3\) is less than 1 and the progression does not converge.
• “\(1 + (x - 2) + (x - 2)^2 + \cdots\) converges for \(x < 3\).” Repair \(-1 < x - 2 < 1\) gives \(1 < x < 3\). A one-sided answer loses the lower bound.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Where this goes next. MF19 prints a second line under Binomial series, for \((1 + x)^n\) when \(n\) is not a positive integer. That series never stops and needs a condition on \(x\); it is Paper 3, chapter 9, and is not used here. \(\binom{n}{r}\) returns in chapter 24 as the number of ways of choosing \(r\) objects from \(n\).
• The condition, as you must be able to state it. A geometric progression converges if and only if \(|r| < 1\) (equivalently \(-1 < r < 1\)), and then \(S_\infty = \dfrac{a}{1-r}\). MF19 prints the condition beside the formula; you must also be able to state it on its own and to use it to find which values of a letter are allowed. Check it before every \(S_\infty\) you write. An arithmetic progression never converges unless every term is zero, so it has no sum to infinity.
• Interleave with the chapters that use this one. When you reach chapter 7, re-expand \((2 + h)^3\) for the chord gradient; at chapter 9, redo worked example 3 before meeting the rational-\(n\) series, so the two expansions stay separate; at chapter 10, redo drill item P9 in lesson card C, the “least \(n\)” for 2, 6, 18, …, with logarithms and compare it with the trial method here; at chapters 24 and 26, recompute \(\binom{7}{3}\) and a sum to infinity. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: Paper 1 is assumed knowledge for every other paper, and an individual examination question may involve ideas and methods from more than one section of that paper’s content, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 1 content, examined in Paper 1. Paper 1 (Pure Mathematics 1) is compulsory for both the AS Level and the A Level: it is 60% of the AS Level and 30% of the A Level, and its content is assumed knowledge for every other paper. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A binomial part asks for an expansion, one coefficient, a term independent of \(x\), or an unknown constant found from a given coefficient, and may carry on into a product of two brackets. A progression part gives two facts and asks for \(a\) and \(d\) (or \(a\) and \(r\)), then a later term, a sum, a least \(n\) or a sum to infinity. The syllabus says questions may involve more than one progression, and a “show that” built on \(2b = a + c\) or \(b^2 = ac\) is one way that happens.
• Printed: \(u_n\) and \(S_n\) for both progressions, \(S_\infty = \dfrac{a}{1-r}\) for \(|r| < 1\), and the expansion of \((a + b)^n\) with \(\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}\). To be known: the meaning of every letter, \(l = a + (n-1)d\), the tests \(2b = a + c\) and \(b^2 = ac\), the general term, \(u_n = S_n - S_{n-1}\), and the convergence condition as a statement of its own. The line MF19 prints under the expansion, for rational \(n\), is Paper 3.
• Coefficients, terms and sums are exact: \(-160\), \(765\), \(\tfrac{32}{3}\), \(\tfrac{3}{11}\), never a rounded decimal. A “least \(n\)” answer is an integer taken above an unrounded critical value and confirmed by \(S_n\) either side. Every coefficient is shown built from \(\binom{n}{r}\), the power of the number and the sign; a bare number from a calculator is an unsupported answer. Money and lengths in a context are given to 3 significant figures at the end.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.