IGCSE and O Level Additional Mathematics: Chapter 11 - Series Study Guide

Bibliographic and Publication Information

• Title: Sue Pemberton Cambridge IGCSE® and O Level Additional Mathematics Coursebook Second edition
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Course Contents Overview

• Introductory Material: Acknowledgements (vi), Introduction (vii), How to use this book (viii).
• Chapter 1: Functions (1) – Mappings, definition of a function, composite functions, modulus functions, graphs of $y = |f(x)|$ linear, inverse functions, functional graphs and inverses.
• Chapter 2: Simultaneous equations and quadratics (23) – Linear/non-linear systems, max/min values of quadratics, graphs of $y = |f(x)|$ quadratic, inequalities, roots, line and curve intersection.
• Chapter 3: Indices and surds (49) – Simplifying indices, equations involving indices, surds, surd operations, rationalising denominators, surd equations.
• Chapter 4: Factors and polynomials (70) – Polynomial operations, division, factor theorem, cubic expressions/equations, remainder theorem.
• Chapter 5: Equations, inequalities and graphs (89) – Modulus equations $|ax - b| = |cx - d|$, modulus inequalities, cubic polynomial sketching, graphical cubic inequalities.
• Chapter 6: Logarithmic and exponential functions (107) – Logs (base 10 and a), laws of logs, log equations, exponential equations, change of base, natural logs, practical applications, graphs, and inverses.
• Chapter 7: Straight-line graphs (134) – Length/mid-point, parallel/perpendicular lines, line equations, rectilinear areas, linear law (non-linear to linear conversion), data relationships.
• Chapter 8: Circular measure (166) – Circular measure basics, arc length, sector area.
• Chapter 9: Trigonometry (182) – Angles $0^{\circ}$ to $90^{\circ}$, general angle definitions, trig ratios, graphs, modulus trig graphs, equations, identities, and advanced equations/identities.
• Chapter 10: Permutations and combinations (220) – Factorial notation, arrangements, permutations, combinations.
• Chapter 11: Series (239) – Pascal’s triangle, binomial theorem, arithmetic progressions, geometric progressions, infinite geometric series.
• Chapter 12: Differentiation 1 (270) – Gradient function, chain rule, product rule, quotient rule, tangents/normals, small increments, rates of change, second derivatives, stationary points, practical max/min problems.
• Chapter 13: Vectors (311) – Vector notation, position vectors, vector geometry, constant velocity.
• Chapter 14: Differentiation 2 (332) – Exponential, logarithmic, and trigonometric derivatives, further applications.
• Chapter 15: Integration (357) – Reversed differentiation, indefinite integrals, specific forms (ax+b)^n, exp and trig integration, $1/x$ form, definite integration, area under curve, area between line and curve.
• Chapter 16: Kinematics (397) – Differentiation and integration applications in kinematics.
• Additional Material: Answers (417), Index (449).

Chapter 11: Series - Learning Objectives

• Use the binomial theorem for expansion of $(a + b)^n$ for positive integral $n$.
• Use the general term $\binom{n}{r} a^{n - r} b^r$ for a binomial expansion.
• Recognize arithmetic and geometric progressions.
• Use formulae for the $n\text{th}$ term and the sum of the first $n$ terms of arithmetic and geometric progressions.
• Use the condition for convergence of a geometric progression and the formula for the sum to infinity of a convergent geometric series.

11.1 Pascal’s Triangle

• Definition of Binomial: The word means 'two terms'. In algebra, it refers to expressions like $x + 5$ or $2x - 3y$.
• Expansion Patterns:
  • $(a + b)^1 = 1a + 1b$
  • $(a + b)^2 = 1a^2 + 2ab + 1b^2$
  • $(a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3$
  • $(a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4$
  • In the expansion of $(a + b)^n$, the power of $a$ decreases by 1 while the power of $b$ increases by 1 in successive terms. All terms have a total index equal to $n$.
• Pascal’s Triangle Construction:
  • Each row starts and ends with 1.
  • Each internal number is the sum of the two numbers directly above it.
  • The 5th row is $1, 5, 10, 10, 5, 1$, corresponding to $(a + b)^5 = 1a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + 1b^5$
• Worked Example 1:
  • Expanding $(2 + 5x)^3$: Using row $1, 3, 3, 1$, result is $1(2)^3 + 3(2)^2(5x) + 3(2)(5x)^2 + 1(5x)^3 = 8 + 60x + 150x^2 + 125x^3$.
  • Expanding $(2x - 3)^4$: Using row $1, 4, 6, 4, 1$, result is $1(2x)^4 + 4(2x)^3(-3) + 6(2x)^2(-3)^2 + 4(2x)(-3)^3 + 1(-3)^4 = 16x^4 - 96x^3 + 216x^2 - 216x + 81$.
• Class Discussion 1:
  • Tetrahedral Numbers: Highlighted sequence $1, 4, 10, 20$ found in Pascal’s triangle.
  • Row Totals: Finding the sum of indices across rows reveals patterns based on powers of 2 (e.g., $1+1=2^1$, $1+2+1=2^2$, etc.).
• Worked Example 2:
  • Expanding $(2 - x)^5 = 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5$.
  • To find the coefficient of $x^3$ in $(1 + 3x)(2 - x)^5$: The $x^3$ term comes from $1 \times (-40x^3)$ and $3x \times 80x^2$. Calculation: $-40 + 240 = 200$.

11.2 The Binomial Theorem

• Formula using Combinations:
  • For positive integer $n$: $(a + b)^n = {}^nC_0 a^n + {}^nC_1 a^{n-1}b + {}^nC_2 a^{n-2}b^2 + \dots + {}^nC_r a^{n-r}b^r + \dots + {}^nC_n b^n$
  • Simplified: $(a + b)^n = a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2}b^2 + \dots + b^n$
• Combination Definition: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
• Worked Example 3: Expanding $(3 + 4x)^5 = 3^5 + 5C1 (3^4)(4x) + 5C2 (3^3)(4x)^2 + 5C3 (3^2)(4x)^3 + 5C4 (3)(4x)^4 + (4x)^5 = 243 + 1620x + 4320x^2 + 5760x^3 + 3840x^4 + 1024x^5$.
• Worked Example 4: Find the coefficient of $x^{20}$ in $(2 - x)^{25}$. The term is ${}^{25}C_{20} \times 2^5 \times (-x)^{20}$. Coefficient = $53130 \times 32 = 1700160$.
• Alternative Binomial Formula:
  • $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \frac{n(n-1)(n-2)(n-3)}{4!}x^4 + \dots$
• Worked Example 5:
  • First 4 terms of $(1+3y)^7$: $1 + 7(3y) + \frac{7 \times 6}{2!}(3y)^2 + \frac{7 \times 6 \times 5}{3!}(3y)^3 = 1 + 21y + 189y^2 + 945y^3$.
  • First 4 terms of $(2 - y)^6 = 2^6 \left[1 + 6\left(-\frac{y}{2}\right) + \frac{6 \times 5}{2!}\left(-\frac{y}{2}\right)^2 + \frac{6 \times 5 \times 4}{3!}\left(-\frac{y}{2}\right)^3\right] = 64 - 192y + 240y^2 - 160y^3$.

11.3 Arithmetic Progressions

• Definition: A number sequence where each term differs from the prior one by a constant called the common difference ($d$).
• Notation:
  • $a$ = first term
  • $d$ = common difference
  • $l$ = last term
• Formula for $n\text{th}$ term: $n\text{th term} = a + (n - 1)d$
• Worked Example 6: Find number of terms in $-17, -14, -11, …, 58$. Using $58 = -17 + 3(n - 1)$, result is $n = 26$.
• Worked Example 7: If $5\text{th term} = 4.4$ and $9\text{th term} = 7.6$. Equations: $a + 4d = 4.4$ and $a + 8d = 7.6$. Solving gives $d = 0.8$, $a = 1.2$.
• Sum of an Arithmetic Progression ($S_n$):
  • Formula 1: $S_n = \frac{n}{2}(a+l)$
  • Formula 2: $S_n = \frac{n}{2}[2a + (n - 1)d]$
• Proof of Sum Formula:
  • Rewrite $S_n$ forwards and backwards:
    • $S_n = a + (a + d) + \dots + (l - d) + l$
    • $S_n = l + (l - d) + \dots + (a + d) + a$
  • Adding these gives $2S_n = n(a + l)$, thus $S_n = \frac{n}{2}(a + l)$.
• Class Discussion (Carl Gauss): At age 8, Gauss found the sum of 1 to 100 by pairing $(1+100), (2+99), \dots$ resulting in $50 \times 101 = 5050$. The method generalizes to $\frac{n(n + 1)}{2}$.
• Worked Example 11: If $S_n = 5n^2 - 3n$. First term $S_1 = 2$. Second term via $S_2 = 14$, so $second term = 12$. Common difference $d = 10$. $n\text{th term} = S_n - S_{n-1} = 10n - 8$.

11.4 Geometric Progressions

• Definition: A sequence where each term is a constant multiple of the previous term. This constant is the common ratio ($r$).
• Notation:
  • $a$ = first term
  • $r$ = common ratio
• Formula for $n\text{th}$ term: $n\text{th term} = ar^{n-1}$
• Worked Example 13: 2nd term is 108, 4th term is 48. $ar = 108$, $ar^3 = 48$. Ratio $r^2 = 48/108 = 4/9$. If terms are positive, $r = 2/3$. First term $a = 108 / (2/3) = 162$.
• Sum of a Geometric Progression ($S_n$):
  • Formula: $S_n = \frac{a(r^n - 1)}{r - 1}$ (use if $r > 1$ or $r < -1$)
  • Formula: $S_n = \frac{a(1 - r^n)}{1 - r}$ (use if $-1 < r < 1$)
  • Note: $\text{Formulae valid for } r \neq 1$.
• Proof of Geometric Sum:
  • $S_n = a + ar + ar^2 + \dots + ar^{n-1}$
  • $rS_n = ar + ar^2 + \dots + ar^{n-1} + ar^n$
  • Subtracting: $rS_n - S_n = ar^n - a \rightarrow S_n(r-1) = a(r^n-1)$.
• Worked Example 15: In sequence $2, 6, 18, …$, which term exceeds $1,000,000$? Solve $2 \times 3^{n-1} > 1,000,000 \rightarrow 3^{n-1} > 500,000$. Using logs: $(n-1) \log(3) > \log(500,000) \rightarrow n-1 > 11.94$. Thus, $n = 13$.

11.5 Infinite Geometric Series

• Convergent Series: A series where the sum to infinity approaches a finite number. This happens if $-1 < r < 1$.
• Sum to Infinity Formula: $S_{\infty} = \frac{a}{1 - r}$.
• Proof Logic: As $n \rightarrow \infty$, $r^n \rightarrow 0$ when $|r| < 1$. The sum formula $S_n = \frac{a(1 - r^n)}{1 - r}$ becomes $\frac{a(1 - 0)}{1 - r}$.
• Worked Example 18: Series $25, 15, 9$. $r = 15/25 = 3/5$. $S_{\infty} = \frac{25}{1 - 0.6} = 62.5$.
• Worked Example 19: $r = -4/5$, $S_4 = 164$. Expand $164 = a(1 - (-4/5)^4) / (1 - (-4/5))$. Solving gives $a = 500$. $S_{\infty} = 500 / (1 + 4/5) = 500 / (9/5) = 277 \text{ } 7/9$.
• Koch Snowflake Challenge:
  • Step 1: Divide segment into 3 parts.
  • Step 2: Draw equilateral triangle on middle segment.
  • Step 3: Remove base.
  • Result: Perimeter diverges to infinity, but area converges to $8/5$ of the original triangle's area.

11.6 Further Arithmetic and Geometric Series

• Hybrid Problems: Some problems mix series types.
• Worked Example 20: Terms $x, y, x^2$ are in arithmetic progression; $x, x^2, y$ are in geometric progression. Case $2y = x^2 + x$ (arithmetic) and $y/x^2 = x^2/x \rightarrow y = x^3$ (geometric).
  • $2x^3 = x^2 + x \rightarrow 2x^2 - x - 1 = 0 \rightarrow (2x+1)(x-1)=0$.
  • Given $x < 0$, then $x = -1/2$ and $y = -1/8$.

Summary of Formulae

• Binomial Expansion: $(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r}b^r$
• Arithmetic Progression:
  • $k\text{th term} = a + (k - 1)d$
  • $S_n = \frac{n}{2}[2a + (n - 1)d]$
• Geometric Progression:
  • $k\text{th term} = ar^{k - 1}$
  • $S_n = \frac{a(1 - r^n)}{1 - r}$
  • $S_{\infty} = \frac{a}{1 - r}$ for $|r| < 1$

Questions & Discussion

• Stepping Stone Game Rules: Moves allowed are East (E) or South (S).
  • Found that moving from START to point Q requires 3 E and 2 S moves. Total routes is equivalent to combinations ${}^5C_3 = 10$.
  • Patterns of routes on the grid match Pascal's Triangle values.
• Exam Example (June 2015):
  • Expansion of $(2 + x^2/2)^6 = 64 + 192x^2 + 240x^4 + 160x^6$
  • Term independent of $x$ in $(2 + x^2/2)^6(1 - 3/x^2 + 9/x^4)$.
  • Calculation: $(64 \times 1) + (192x^2 \times -3/x^2) + (240x^4 \times 9/x^4) = 64 - 576 + 2160 = 1648$.
  • Note variation in source: The textbook example on p36 lists a calculation resulting in 1072 for a slightly different problem form: independent term of $(2 + x^2/2)^6(1 - 6/x^2 + 9/x^4)$.
• Gauss Method Applications: Used to find sums of sequences with different step sizes, e.g., $2 + 4 + 6 + \dots + 400$ or multiples of 3 ($3 + 6 + 9 + \dots$).