IB Math AA HL — Number & Algebra

Complete Study Guide

Topics Covered

1. Sequences & Series — arithmetic and geometric, sigma notation, infinite sums
2. Exponents & Logarithms — laws, change of base, natural log, solving equations
3. Binomial Theorem — expansion, general term, finding specific terms
4. Proof by Induction HL — sum formulas, divisibility, formal structure
5. Systems of Equations HL — 3×3 systems, Gaussian elimination, geometric interpretation
6. Practice Questions & Exam Alerts

Topic 1 of the IB Math AA HL syllabus — Paper 1 and Paper 2

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Section 1: Sequences & Series

A sequence is an ordered list of numbers following a rule. A series is the sum of the terms of a sequence. On the IB exam, sequences and series questions appear on both Paper 1 (no calculator) and Paper 2, and they frequently combine with logarithms or proofs in multi-part questions.

1.1 Arithmetic Sequences & Series

In an arithmetic sequence, each term is obtained by adding a fixed number $d$ (the common difference) to the previous term.

$u_n = u_1 + (n-1)d$

The sum of the first $n$ terms of an arithmetic series:

$S_n = \frac{n}{2}(2u_1 + (n-1)d) \qquad \text{or equivalently} \qquad S_n = \frac{n}{2}(u_1 + u_n)$

The second form is often faster when you already know the first and last terms.

1.2 Geometric Sequences & Series

In a geometric sequence, each term is obtained by multiplying the previous term by a fixed number $r$ (the common ratio):

$u_n = u_1 \cdot r^{n-1}$

Sum of the first $n$ terms (for $r \neq 1$):

$S_n = \frac{u_1(1 - r^n)}{1 - r}$

An equivalent form sometimes used when $r > 1$:

$S_n = \frac{u_1(r^n - 1)}{r - 1}$

Infinite geometric series: When $\lvert r \rvert < 1$, the partial sums converge:

$S_\infty = \frac{u_1}{1 - r}, \qquad \lvert r \rvert < 1$

1.3 Sigma Notation

Sigma notation provides a compact way to write series:

$\sum_{k=1}^{n} u_k = u_1 + u_2 + \cdots + u_n$

Key properties:

$\sum_{k=1}^{n} (a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \qquad \sum_{k=1}^{n} c \cdot a_k = c \sum_{k=1}^{n} a_k$

$\sum_{k=1}^{n} c = nc \qquad \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

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Section 2: Exponents & Logarithms

2.1 Laws of Exponents

For $a > 0$, $b > 0$, and $m, n \in \mathbb{R}$:

Law	Expression
Product	$a^m \cdot a^n = a^{m+n}$
Quotient	$\dfrac{a^m}{a^n} = a^{m-n}$
Power of power	$(a^m)^n = a^{mn}$
Negative exponent	$a^{-n} = \dfrac{1}{a^n}$
Fractional exponent	$a^{m/n} = \sqrt[n]{a^m}$

2.2 Laws of Logarithms

For $a > 0$, $a \neq 1$, $x > 0$, $y > 0$:

$\log_a(xy) = \log_a x + \log_a y$

$\log_a\!\left(\frac{x}{y}\right) = \log_a x - \log_a y$

$\log_a(x^n) = n \log_a x$

$\log_a a = 1 \qquad \log_a 1 = 0$

Change of base:

$\log_a x = \frac{\log_b x}{\log_b a} \qquad \text{(commonly: } \log_a x = \frac{\ln x}{\ln a}\text{)}$

The natural logarithm $\ln x = \log_e x$ where $e \approx 2.718$. On IB exams, $\ln$ and $e^x$ are inverse functions: $e^{\ln x} = x$ and $\ln(e^x) = x$.

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Section 3: Binomial Theorem

3.1 The Expansion

The binomial theorem states that for $n \in \mathbb{Z}^+$:

$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$

where the binomial coefficient is:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

Pascal’s triangle provides $\binom{n}{r}$ values for small $n$, but for larger $n$ you should use the formula directly or the GDC.

3.2 The General Term

The $(r+1)$th term (also written $T_{r+1}$) of the expansion of $(a+b)^n$ is:

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Note the indexing: $r$ starts at 0, so the first term is $T_1$ (with $r=0$) and the last term is $T_{n+1}$ (with $r=n$).

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Section 4: Proof by Induction HL

Proof by mathematical induction is an HL-only technique and is consistently tested on Paper 1. It proves that a statement $P(n)$ holds for all integers $n \geq n_0$ (usually $n_0 = 1$).

4.1 The Formal Structure

Every induction proof must contain all four components:

1. Base case: Show $P(n_0)$ is true by direct calculation.
2. Inductive hypothesis: State “Assume $P(k)$ is true for some $k \geq n_0$”, then write out explicitly what that assumption says.
3. Inductive step: Using the hypothesis, prove that $P(k+1)$ is true.
4. Conclusion: State “Therefore, by the principle of mathematical induction, $P(n)$ is true for all $n \geq n_0$.”

4.2 Worked Example: Sum of a Series

4.3 Worked Example: Divisibility Proof

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Section 5: Systems of Equations HL

5.1 Setting Up a 3×3 System

A system of three linear equations in three unknowns $x$, $y$, $z$ can be written in augmented matrix form:

$\begin{pmatrix} a_1 & b_1 & c_1 & \mid & d_1 \\ a_2 & b_2 & c_2 & \mid & d_2 \\ a_3 & b_3 & c_3 & \mid & d_3 \end{pmatrix}$

5.2 Gaussian Elimination

The goal is to reduce the augmented matrix to row echelon form (upper triangular) using three elementary row operations (EROs):

Operation	Notation	Description
Swap rows	$R_i \leftrightarrow R_j$	Interchange two rows
Scale a row	$kR_i \to R_i$	Multiply a row by a non-zero constant
Add multiple	$R_i + kR_j \to R_i$	Add a multiple of one row to another

Once in row echelon form, solve by back substitution from the bottom row up.

5.3 Geometric Interpretation

Each equation $ax + by + cz = d$ represents a plane in three-dimensional space. A $3 \times 3$ system describes the intersection of three planes:

Reduced form outcome	Geometric meaning	Number of solutions
Full diagonal (no zeros)	Three planes meet at a single point	Unique solution
Row of zeros, consistent	Planes share a line or a plane	Infinitely many
Row $0\,0\,0\,\mid\,c$, $c \neq 0$	Planes are inconsistent	No solution

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Section 6: Practice Questions

Question 1 — Arithmetic series (Paper 1 style)

The sum of the first $n$ terms of an arithmetic series is $S_n = 3n^2 + 2n$.

(a) Find the first term and the common difference.

(b) Find the 15th term.

Show full solution

(a) First term: $u_1 = S_1 = 3(1)^2 + 2(1) = 5$.

Second term: $u_2 = S_2 - S_1 = (12 + 4) - 5 = 11$.

Common difference: $d = u_2 - u_1 = 11 - 5 = 6$.

(b) Using $u_n = u_1 + (n-1)d$:

$u_{15} = 5 + 14 \times 6 = 5 + 84 = 89$

Alternatively, $u_{15} = S_{15} - S_{14} = (675 + 30) - (588 + 28) = 705 - 616 = 89$. ✓

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Question 2 — Infinite geometric series

A geometric series has $u_1 = 8$ and $u_3 = 2$.

(a) Find the two possible values of the common ratio.

(b) For the value of $r$ for which the series converges, find $S_\infty$.

(c) Find the smallest value of $n$ such that $S_n > 0.99 \, S_\infty$.

Show full solution

(a) $u_3 = u_1 r^2 \Rightarrow 2 = 8r^2 \Rightarrow r^2 = \tfrac{1}{4} \Rightarrow r = \pm\tfrac{1}{2}$.

(b) Both $r = \tfrac{1}{2}$ and $r = -\tfrac{1}{2}$ satisfy $\lvert r \rvert < 1$, so both converge. For $r = \tfrac{1}{2}$:

$S_\infty = \frac{8}{1 - \tfrac{1}{2}} = \frac{8}{0.5} = 16$

For $r = -\tfrac{1}{2}$:

$S_\infty = \frac{8}{1 + \tfrac{1}{2}} = \frac{8}{1.5} = \frac{16}{3}$

(If the question specifies positive terms, take $r = \tfrac{1}{2}$, giving $S_\infty = 16$.)

(c) For $r = \tfrac{1}{2}$, $S_\infty = 16$. We need $S_n > 0.99 \times 16 = 15.84$:

$S_n = 16\!\left(1 - \left(\tfrac{1}{2}\right)^n\right) > 15.84$

$1 - \left(\tfrac{1}{2}\right)^n > 0.99 \implies \left(\tfrac{1}{2}\right)^n < 0.01$

Take logarithms: $n \ln\tfrac{1}{2} < \ln 0.01 \Rightarrow n > \dfrac{\ln 0.01}{\ln 0.5} = \dfrac{-4.605}{-0.693} \approx 6.64$.

Since $n$ must be a positive integer, the smallest value is $n = 7$.

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Question 3 — Binomial theorem

(a) Expand $(1 + 2x)^5$ in ascending powers of $x$ up to and including the term in $x^3$.

(b) Find the coefficient of $x^4$ in the expansion of $(3 - x)^7$.

(c) HL Find the value of $k$ such that the expansion of $(k + x)^6$ has a term in $x^2$ equal to $60x^2$.

Show full solution

(a) Using $T_{r+1} = \binom{5}{r}(1)^{5-r}(2x)^r = \binom{5}{r} 2^r x^r$:

$r$	$\binom{5}{r}$	$2^r$	Term
0	1	1	$1$
1	5	2	$10x$
2	10	4	$40x^2$
3	10	8	$80x^3$

$(1+2x)^5 = 1 + 10x + 40x^2 + 80x^3 + \cdots$

(b) General term of $(3-x)^7$: $T_{r+1} = \binom{7}{r}3^{7-r}(-x)^r = \binom{7}{r}3^{7-r}(-1)^r x^r$.

For $x^4$: $r = 4$.

$T_5 = \binom{7}{4} 3^3 (-1)^4 x^4 = 35 \times 27 \times 1 \times x^4 = 945\,x^4$

Coefficient is $945$.

(c) General term of $(k+x)^6$: $T_{r+1} = \binom{6}{r} k^{6-r} x^r$.

For $x^2$: $r = 2$.

$T_3 = \binom{6}{2} k^4 x^2 = 15k^4 x^2$

Set equal to $60x^2$: $15k^4 = 60 \Rightarrow k^4 = 4 \Rightarrow k = \pm\sqrt{2}$.

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Question 4 — Proof by induction HL

Prove by mathematical induction that $\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all $n \in \mathbb{Z}^+$.

Show full solution

State $P(n)$: $\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$.

Base case ($n=1$):

LHS $= 1^2 = 1$. RHS $= \dfrac{1 \times 2 \times 3}{6} = 1$.

LHS $=$ RHS. $P(1)$ is true. $\checkmark$

Inductive hypothesis: Assume $P(k)$ is true for some $k \in \mathbb{Z}^+$:

$\sum_{r=1}^{k} r^2 = \frac{k(k+1)(2k+1)}{6}$

Inductive step: Show $P(k+1)$: $\displaystyle\sum_{r=1}^{k+1} r^2 = \dfrac{(k+1)(k+2)(2k+3)}{6}$.

$\sum_{r=1}^{k+1} r^2 = \sum_{r=1}^{k} r^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$

Factor out $(k+1)$:

$= (k+1)\!\left[\frac{k(2k+1)}{6} + (k+1)\right] = (k+1) \cdot \frac{k(2k+1) + 6(k+1)}{6}$

Expand the numerator:

$k(2k+1) + 6(k+1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6 = (k+2)(2k+3)$

Therefore:

$\sum_{r=1}^{k+1} r^2 = \frac{(k+1)(k+2)(2k+3)}{6}$

This is the RHS of $P(k+1)$. $\checkmark$

Conclusion: Since $P(1)$ is true, and $P(k) \Rightarrow P(k+1)$ for all $k \in \mathbb{Z}^+$, by the principle of mathematical induction $P(n)$ is true for all $n \in \mathbb{Z}^+$. $\blacksquare$

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Question 5 — Systems of equations HL

Consider the system of equations:

$x + 2y + z = 3$ $2x + y + az = b$ $x - y + 2z = 1$

(a) Show that when $a = 3$ and $b = 4$, the system has infinitely many solutions.

(b) Find the condition on $a$ and $b$ under which the system has a unique solution.

Show full solution

Augmented matrix:

$\left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 2 & 1 & a & b \\ 1 & -1 & 2 & 1 \end{array}\right)$

$R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - R_1$:

$\left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & -3 & a-2 & b-6 \\ 0 & -3 & 1 & -2 \end{array}\right)$

$R_3 \to R_3 - R_2$:

$\left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & -3 & a-2 & b-6 \\ 0 & 0 & 3-a & -2-(b-6) \end{array}\right) = \left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & -3 & a-2 & b-6 \\ 0 & 0 & 3-a & 4-b \end{array}\right)$

(a) When $a = 3$, $b = 4$: bottom row is $(0\ 0\ 0\ \mid\ 0)$, i.e., $0 = 0$. Infinitely many solutions. $\checkmark$

(b) Unique solution requires the bottom-right pivot to be non-zero: $3 - a \neq 0$, i.e., $a \neq 3$.

When $a \neq 3$, the system has a unique solution for any value of $b$.

(When $a = 3$ and $b \neq 4$, the bottom row gives $0 = 4 - b \neq 0$: no solution.)

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Question 6 — Mixed: logarithms and series

The $n$th term of a sequence is $u_n = \log_2(2^n \cdot 3)$. Show that the sequence is arithmetic and find its common difference.

Show full solution

Use the log product law: $u_n = \log_2(2^n) + \log_2 3 = n + \log_2 3$.

This is linear in $n$: $u_n = n + \log_2 3$.

Common difference: $d = u_{n+1} - u_n = (n+1+\log_2 3) - (n + \log_2 3) = 1$.

Since $d$ is constant, the sequence is arithmetic with common difference $d = 1$. $\blacksquare$

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