Series

1. Overview

The topic of Series in IGCSE Additional Mathematics (0606) focuses on recognising and manipulating arithmetic and geometric progressions, expanding binomial expressions, and understanding the concept of convergence. These skills are crucial for solving problems involving patterns, growth, and approximations. Expect direct questions on AP/GP terms and sums, binomial expansions (often linked to finding specific coefficients), and determining if a geometric series has a sum to infinity. Mastery of series is essential not only for scoring well on these specific questions but also for building a foundation for more advanced mathematical concepts.

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Key Definitions

• Sequence: An ordered list of numbers following a specific rule.
• Series: The sum of the terms of a sequence.
• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant (the common difference, $d$).
• Geometric Progression (GP): A sequence where each term is found by multiplying the previous term by a constant (the common ratio, $r$).
• Convergent Series: A geometric series where the sum of terms approaches a finite limit as the number of terms increases to infinity (occurs when $|r| < 1$).
• Binomial: An algebraic expression consisting of two terms, e.g., $(a + b)$.

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Core Content

A. The Binomial Theorem

For any positive integer $n$, the expansion of $(a + b)^n$ is given by: $$(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{r}a^{n-r}b^r + \dots + b^n$$ Note: This formula is provided on the formula sheet.

The General Term:
To find a specific term without expanding the whole bracket, use: $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

• Remember that the $(r+1)^{th}$ term uses $r$ in the combination formula (e.g., the 4th term uses $r=3$).
• The coefficients $\binom{n}{r}$ can be found using the formula $\frac{n!}{r!(n-r)!}$ or Pascal's Triangle.

Worked example 1 — Full Expansion

Find the expansion of $(2 - x)^4$ in ascending powers of $x$.

Step 1: Identify $a$, $b$, and $n$
$a = 2, b = (-x), n = 4$.

Step 2: Apply the binomial theorem
$(2-x)^4 = \binom{4}{0}(2)^4(-x)^0 + \binom{4}{1}(2)^3(-x)^1 + \binom{4}{2}(2)^2(-x)^2 + \binom{4}{3}(2)^1(-x)^3 + \binom{4}{4}(2)^0(-x)^4$

Step 3: Evaluate the binomial coefficients
$(2-x)^4 = (1)(2)^4(-x)^0 + (4)(2)^3(-x)^1 + (6)(2)^2(-x)^2 + (4)(2)^1(-x)^3 + (1)(2)^0(-x)^4$

Step 4: Simplify each term
$(2-x)^4 = 1(16)(1) + 4(8)(-x) + 6(4)(x^2) + 4(2)(-x^3) + 1(1)(x^4)$
Simplify powers and products

Step 5: Final simplification
$(2-x)^4 = 16 - 32x + 24x^2 - 8x^3 + x^4$
Combine the terms to get the final expansion

Final Answer:
$(2-x)^4 = \boxed{16 - 32x + 24x^2 - 8x^3 + x^4}$

Worked example 2 — Finding a Specific Coefficient

Find the coefficient of $x^2$ in the expansion of $(3 + 2x)(1 - \frac{x}{2})^6$.

Step 1: Expand $(1 - \frac{x}{2})^6$ up to the $x^2$ term
$(1 - \frac{x}{2})^6 \approx \binom{6}{0}(1)^6 + \binom{6}{1}(1)^5(-\frac{x}{2}) + \binom{6}{2}(1)^4(-\frac{x}{2})^2$
Apply the binomial theorem, only expanding up to the $x^2$ term as higher powers won't contribute to the $x^2$ term in the final product.

Step 2: Evaluate the binomial coefficients and simplify
$(1 - \frac{x}{2})^6 \approx 1 + 6(-\frac{x}{2}) + 15(\frac{x^2}{4})$

Step 3: Simplify further
$(1 - \frac{x}{2})^6 \approx 1 - 3x + \frac{15}{4}x^2$
Simplify powers and products

Step 4: Multiply by $(3 + 2x)$ and identify terms resulting in $x^2$
$(3 + 2x)(1 - 3x + \frac{15}{4}x^2) = 3(1 - 3x + \frac{15}{4}x^2) + 2x(1 - 3x + \frac{15}{4}x^2)$
Distribute the terms

Step 5: Identify the relevant $x^2$ terms
$3 \times \frac{15}{4}x^2 + 2x \times -3x = \frac{45}{4}x^2 - 6x^2$
Only these two terms will result in $x^2$ after multiplication

Step 6: Combine the $x^2$ terms
$(\frac{45}{4} - 6)x^2 = (\frac{45}{4} - \frac{24}{4})x^2 = \frac{21}{4}x^2$

Final Answer:
The coefficient of $x^2$ is $\boxed{\frac{21}{4}}$

B. Arithmetic Progressions (AP)

An AP follows the pattern: $a, a+d, a+2d, \dots$

• $n^{th}$ term: $u_n = a + (n-1)d$
• Sum of first $n$ terms: $S_n = \frac{n}{2}(2a + (n-1)d)$ (F) or $S_n = \frac{n}{2}(a + l)$, where $l$ is the last term.

C. Geometric Progressions (GP)

A GP follows the pattern: $a, ar, ar^2, \dots$

• $n^{th}$ term: $u_n = ar^{n-1}$
• Sum of first $n$ terms: $S_n = \frac{a(1-r^n)}{1-r}$ (F) or $S_n = \frac{a(r^n-1)}{r-1}$ (where $r \neq 1$)
• Sum to Infinity ($S_\infty$): Only exists if the series is convergent, meaning $|r| < 1$ (or $-1 < r < 1$). $$S_\infty = \frac{a}{1-r}$$ (F)

Worked example 3 — GP and Convergence

The second term of a GP is 8 and the sum to infinity is 32. Find the possible values of the common ratio $r$.

Step 1: Set up equations
$u_2 = ar = 8$
The second term of a GP is given by $ar$

$S_\infty = \frac{a}{1-r} = 32$
The sum to infinity of a GP is given by $\frac{a}{1-r}$

Step 2: Express $a$ in terms of $r$ from the first equation
$a = \frac{8}{r}$
Divide both sides of the equation $ar = 8$ by $r$

Step 3: Substitute $a$ into the $S_\infty$ formula
$\frac{8/r}{1-r} = 32$

Step 4: Simplify the equation
$\frac{8}{r(1-r)} = 32$

Step 5: Rearrange the equation
$8 = 32r(1-r)$
Multiply both sides by $r(1-r)$

Step 6: Expand and rearrange into a quadratic equation
$8 = 32r - 32r^2$
Expand the right side

$32r^2 - 32r + 8 = 0$
Rearrange to get a quadratic equation in the form $ar^2 + br + c = 0$

Step 7: Simplify the quadratic equation by dividing by 8
$4r^2 - 4r + 1 = 0$

Step 8: Solve the quadratic equation
$(2r - 1)^2 = 0$
Factorise the quadratic equation

$2r - 1 = 0$
Take the square root of both sides

$r = \frac{1}{2}$
Solve for $r$

Step 9: Check the condition for the sum to infinity
Since $|\frac{1}{2}| < 1$, the sum to infinity exists.
The sum to infinity exists only if $|r| < 1$

Final Answer:
$r = \boxed{\frac{1}{2}}$

Worked example 4 — Arithmetic Progression

The sum of the first 5 terms of an arithmetic progression is 40, and the 8th term is 25. Find the first term and the common difference.

Step 1: Write down the formula for the sum of the first $n$ terms of an AP
$S_n = \frac{n}{2}(2a + (n-1)d)$
This formula is provided on the formula sheet

Step 2: Apply the formula for $S_5 = 40$
$40 = \frac{5}{2}(2a + (5-1)d)$

Step 3: Simplify the equation
$40 = \frac{5}{2}(2a + 4d)$
Simplify inside the brackets

$80 = 5(2a + 4d)$
Multiply both sides by 2

$16 = 2a + 4d$
Divide both sides by 5

$8 = a + 2d$ (Equation 1)
Divide both sides by 2

Step 4: Write down the formula for the $n$th term of an AP
$u_n = a + (n-1)d$

Step 5: Apply the formula for $u_8 = 25$
$25 = a + (8-1)d$

$25 = a + 7d$ (Equation 2)
Simplify inside the brackets

Step 6: Solve the simultaneous equations
Subtract Equation 1 from Equation 2:
$25 - 8 = (a + 7d) - (a + 2d)$
Eliminate $a$ by subtracting the equations

$17 = 5d$
Simplify

$d = \frac{17}{5}$
Solve for $d$

Step 7: Substitute $d$ back into Equation 1 to find $a$
$8 = a + 2(\frac{17}{5})$

$8 = a + \frac{34}{5}$
Simplify

$a = 8 - \frac{34}{5}$
Rearrange to solve for $a$

$a = \frac{40}{5} - \frac{34}{5}$
Express 8 as a fraction with a denominator of 5

$a = \frac{6}{5}$
Simplify

Final Answer:
$a = \boxed{\frac{6}{5}}$, $d = \boxed{\frac{17}{5}}$

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In More Depth

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations

(F) = Provided on the formula sheet

Application	Formula	Variables
Binomial Expansion	$(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r}b^r$ (F)	$n$: power, $r$: term index
AP $n$th term	$u_n = a + (n-1)d$	$a$: 1st term, $d$: difference
AP Sum	$S_n = \frac{n}{2}(2a + (n-1)d)$ (F)	$n$: number of terms
GP $n$th term	$u_n = ar^{n-1}$	$a$: 1st term, $r$: ratio
GP Sum	$S_n = \frac{a(1-r^n)}{1-r}$ (F)	$r \neq 1$
Sum to Infinity	$S_\infty = \frac{a}{1-r}$ (F)	Condition: $|r| < 1$

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Common Mistakes to Avoid

• ❌ Wrong: Expanding $(3x)^2$ as $3x^2$.
  ✓ Right: $(3x)^2 = 9x^2$. Always apply the power to the coefficient inside the bracket.
• ❌ Wrong: Using the Sum to Infinity formula for a series where $r = 1.5$.
  ✓ Right: Always check if $|r| < 1$ before using $S_\infty$. If $|r| \geq 1$, state that the sum to infinity does not exist because the series diverges.
• ❌ Wrong: Confusing $n$ with $r$ in the binomial general term.
  ✓ Right: In the term $\binom{n}{r}$, $n$ is the power of the bracket and $r$ is one less than the term number (for the $4^{th}$ term, $r=3$).
• ❌ Wrong: Subtracting $S_{10}$ from $S_{30}$ to find the sum of terms from 10 to 30.
  ✓ Right: To find the sum from the $11^{th}$ to the $30^{th}$ term, calculate $S_{30} - S_{10}$.
• ❌ Wrong: Giving a decimal approximation when an exact value (surd, fraction, ln) is required. For example, writing 0.33 instead of $\frac{1}{3}$.
  ✓ Right: Unless explicitly told to round, always provide answers in their exact form (fractions, surds, or in terms of $\ln$).
• ❌ Wrong: Forgetting the negative sign when substituting into the binomial theorem.
  ✓ Right: When $b$ is negative, e.g. $(2-x)^n$, remember to include the negative sign: $b = -x$. Pay close attention to signs when raising negative terms to powers.
• ❌ Wrong: Not simplifying binomial coefficients correctly, especially in Paper 1 (non-calculator).
  ✓ Right: Practice calculating binomial coefficients by hand. Remember $\binom{n}{0} = \binom{n}{n} = 1$ and $\binom{n}{1} = n$.
• ❌ Wrong: Not checking the condition $|r| < 1$ when using the sum to infinity formula.
  ✓ Right: Always verify that the absolute value of the common ratio is less than 1 before applying the formula $S_\infty = \frac{a}{1-r}$. If $|r| \ge 1$, the series does not converge, and the sum to infinity does not exist.

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Exam Tips

• Command Words: If a question says "Show that," every algebraic step must be shown clearly. If you skip steps, you will lose "method marks" (M marks).
• Paper 1 (Non-Calculator): Expect to work with surds and fractions. Do not convert $\frac{1}{3}$ to 0.33. Keep $\sqrt{2}$ as $\sqrt{2}$.
• Checking Restrictions: In Paper 2, if you solve a quadratic for $r$ and get $r=2$ and $r=0.5$, but the question mentions a "sum to infinity," you must reject $r=2$ and explain why ($|r| < 1$ for convergence).
• Efficiency: Before starting a long expansion, check if the question only asks for a specific coefficient. Use the general term formula $\binom{n}{r} a^{n-r} b^r$ to save time.
• Units and Notation: Always state your formulas before substituting values. This ensures you pick up marks even if your arithmetic fails.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [9 marks]

Question:

(a) The first three terms of an arithmetic progression are $\log_3 x$, $\log_3 9x$, and $\log_3 81x$. Find the common difference of the progression. [3]

(b) The $n$th term of a geometric progression is given by $ar^{n-1}$. The 2nd term of a geometric progression is 10 and the 4th term is $\frac{5}{2}$. Find the first term, $a$, and the common ratio, $r$, of the progression. [6]

Worked Solution:

(a)

1. Find the common difference $d$ by subtracting consecutive terms.
   $d = \log_3 9x - \log_3 x = \log_3 \frac{9x}{x} = \log_3 9 = 2$
   $d = \log_3 81x - \log_3 9x = \log_3 \frac{81x}{9x} = \log_3 9 = 2$

2. State the common difference.
   $\boxed{d = 2}$

How to earn full marks: Show your working clearly, demonstrating the logarithmic subtraction and simplification to arrive at the common difference.

(b)

1. Write down the equations for the 2nd and 4th terms.
   $ar = 10$ (1)
   $ar^3 = \frac{5}{2}$ (2)
   Using the formula for the nth term of a geometric progression

2. Divide equation (2) by equation (1).
   $\frac{ar^3}{ar} = \frac{5/2}{10}$
   $r^2 = \frac{1}{4}$
   Dividing the two equations to eliminate $a$

3. Solve for $r$. Since the problem does not specify positive $r$, we must consider both roots.
   $r = \pm \frac{1}{2}$

4. Substitute each value of $r$ into equation (1) to find $a$.
   If $r = \frac{1}{2}$, then $a(\frac{1}{2}) = 10$, so $a = 20$
   If $r = -\frac{1}{2}$, then $a(-\frac{1}{2}) = 10$, so $a = -20$

5. State the two possible solutions for $a$ and $r$.
   $\boxed{a = 20, r = \frac{1}{2}}$ or $\boxed{a = -20, r = -\frac{1}{2}}$

How to earn full marks: Remember to find both possible solutions for a and r, including the negative root, and clearly state them as pairs.

Common Pitfall: Remember that when solving for the common ratio, $r$, in a geometric progression, you need to consider both positive and negative roots unless the question specifies otherwise. Forgetting the negative root will lead to an incomplete solution.

Exam-Style Question 2 — Paper 2 (Calculator Allowed) [7 marks]

Question:

The sum of the first $n$ terms of an arithmetic series is given by $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ is the first term and $d$ is the common difference.

(a) The sum of the first 15 terms of an arithmetic series is 345. The first term is 3. Find the common difference, $d$. [3]

(b) The first term of a geometric series is 20 and the common ratio is 0.7. Find the sum to infinity of this series. [2]

(c) Explain why the arithmetic series in part (a) does not have a sum to infinity. [2]

Worked Solution:

(a)

1. Substitute the given values into the formula for $S_n$.
   $345 = \frac{15}{2}[2(3) + (15-1)d]$
   $345 = \frac{15}{2}[6 + 14d]$

2. Simplify and solve for $d$.
   $345 = 45 + 105d$
   $300 = 105d$
   $d = \frac{300}{105} = \frac{20}{7}$

3. State the value of $d$.
   $\boxed{d = \frac{20}{7}}$ or $\boxed{d = 2.86}$ (3 s.f.)

How to earn full marks: Show the substitution into the formula and all steps in solving for d, giving the answer as a fraction or decimal to 3 s.f.

(b)

1. Use the formula for the sum to infinity of a geometric series.
   $S_\infty = \frac{a}{1-r} = \frac{20}{1-0.7}$

2. Calculate the sum to infinity.
   $S_\infty = \frac{20}{0.3} = \frac{200}{3}$

3. State the sum to infinity.
   $\boxed{S_\infty = \frac{200}{3}}$ or $\boxed{S_\infty = 66.7}$ (3 s.f.)

How to earn full marks: Write down the formula for the sum to infinity, substitute correctly, and give your answer as a fraction or decimal to 3 s.f.

(c)

1. Explain the condition for a sum to infinity in an arithmetic series.
   Arithmetic series do not have a sum to infinity because they do not converge. The terms either increase indefinitely or decrease indefinitely (unless the common difference is zero), so the sum will tend towards infinity (or negative infinity).

How to earn full marks: Explain that arithmetic series do not converge (unless d = 0) and therefore the sum tends to infinity.

Common Pitfall: Make sure you understand the difference between arithmetic and geometric series. Using the wrong formula will lead to an incorrect answer. Also, remember that an arithmetic series only has a sum to infinity if the common difference is zero.

Exam-Style Question 3 — Paper 1 (No Calculator Allowed) [11 marks]

Question:

(a) The binomial expansion of $(3+x)^4$ is $A + Bx + Cx^2 + Dx^3 + Ex^4$. Find the values of the coefficients $A$, $B$, and $C$. [6]

(b) Find the term independent of $x$ in the expansion of $\left(x^3 + \frac{1}{x}\right)^8$. [5]

Worked Solution:

(a)

1. Use the binomial theorem to expand $(3+x)^4$.
   $(3+x)^4 = \binom{4}{0}3^4x^0 + \binom{4}{1}3^3x^1 + \binom{4}{2}3^2x^2 + \binom{4}{3}3^1x^3 + \binom{4}{4}x^4$

2. Calculate the binomial coefficients and simplify.
   $= 1(81)(1) + 4(27)x + 6(9)x^2 + 4(3)x^3 + 1(1)x^4$
   $= 81 + 108x + 54x^2 + 12x^3 + x^4$

3. Identify the coefficients $A$, $B$, and $C$.
   $A = 81$, $B = 108$, $C = 54$

4. State the values of $A$, $B$, and $C$.
   $\boxed{A=81}$, $\boxed{B=108}$, $\boxed{C=54}$

How to earn full marks: Show the full binomial expansion, including the binomial coefficients, and clearly identify the values of A, B, and C.

(b)

1. Write down the general term in the binomial expansion of $\left(x^3 + \frac{1}{x}\right)^8$.
   $T_{r+1} = \binom{8}{r} (x^3)^{8-r} \left(\frac{1}{x}\right)^r = \binom{8}{r} x^{24-3r} x^{-r} = \binom{8}{r} x^{24-4r}$

2. Find the value of $r$ such that the power of $x$ is 0.
   $24 - 4r = 0$
   $4r = 24$
   $r = 6$

3. Substitute $r=6$ into the general term.
   $T_{6+1} = \binom{8}{6} x^{24-4(6)} = \binom{8}{6} x^0 = \binom{8}{6}$
   Finding the term independent of x

4. Calculate the binomial coefficient.
   $\binom{8}{6} = \frac{8!}{6!2!} = \frac{8 \times 7}{2 \times 1} = 28$
   Calculating the final answer

5. State the term independent of $x$.
   $\boxed{28}$

How to earn full marks: Find the general term, equate the power of x to zero to find r, then calculate the binomial coefficient for that value of r.

Common Pitfall: When expanding binomials, be very careful with the powers and coefficients. A small mistake in calculating the binomial coefficients or simplifying the powers can lead to a completely wrong answer. Double-check your work, especially when dealing with larger powers.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [9 marks]

Question:

A charity is organizing a fundraising campaign. They plan to collect donations each day for 25 days.

(a) On the first day, they collect $20. On each subsequent day, they increase the amount collected by $5. Find the total amount collected over the 25 days. [3]

(b) Alternatively, they could collect $20 on the first day, and then increase the amount collected each day by 6%. Find the total amount collected over the 25 days using this alternative plan, giving your answer to the nearest cent. [4]

(c) If the charity aims to collect at least $2000, explain which plan, (a) or (b), they should use. [2]

Worked Solution:

(a)

1. Identify the arithmetic series parameters: $a = 20$, $d = 5$, $n = 25$.

2. Use the formula for the sum of an arithmetic series: $S_n = \frac{n}{2}[2a + (n-1)d]$.
   $S_{25} = \frac{25}{2}[2(20) + (25-1)(5)]$

3. Calculate the sum.
   $S_{25} = \frac{25}{2}[40 + 120] = \frac{25}{2}[160] = 25[80] = 2000$

4. State the total amount.
   $\boxed{2000}$

How to earn full marks: Correctly identify the series as arithmetic, state the formula, substitute the values, and calculate the sum accurately.

(b)

1. Identify the geometric series parameters: $a = 20$, $r = 1.06$, $n = 25$.

2. Use the formula for the sum of a geometric series: $S_n = \frac{a(r^n - 1)}{r - 1}$.
   $S_{25} = \frac{20(1.06^{25} - 1)}{1.06 - 1}$

3. Calculate the sum.
   $S_{25} = \frac{20(4.29187063 - 1)}{0.06} = \frac{20(3.29187063)}{0.06} = 1097.29021$

4. State the total amount to the nearest cent.
   $\boxed{1097.29}$

How to earn full marks: Correctly identify the series as geometric, state the formula, substitute the values, and round your final answer to the nearest cent.

(c)

1. Compare the total amount collected for each plan to the target amount.
   Plan (a) collects $2000, and Plan (b) collects $1097.29.

2. Explain which plan the charity should use.
   The charity aims to collect at least $2000. Plan (a) collects exactly $2000, while plan (b) collects less than $2000. Therefore, the charity should use plan (a) to meet their target.

How to earn full marks: State the amounts collected by each plan and clearly explain why plan (a) is the better choice to meet the charity's target.

Common Pitfall: Make sure you correctly identify whether a problem involves an arithmetic or geometric series. Applying the wrong formula will result in an incorrect answer. Also, pay attention to the required degree of accuracy when giving your final answer.