The theory · Mathematics: analysis and approaches

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Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 1 — Number and algebra

Topic 1 is where the course tests whether you can be exact. Almost everything in it has a formula in the booklet, which means the marks are never for remembering the formula — they are for choosing the right one, substituting carefully, and writing the answer in the form asked for.

This page covers standard level in full and flags the higher level extensions. Read it once through, then use the worked examples.

1. Sequences and series

A sequence is a list of terms, u₁, u₂, u₃, …. A series is what you get when you add them up. Sₙ always means the sum of the first n terms.

Confusing uₙ with Sₙ is the single most common way to lose marks in this topic. u₅ is the fifth term on its own; S₅ is the first five terms added together. Read which one the question wants before you reach for a formula.

Arithmetic — a constant difference

Each term is the previous one plus a fixed number d, the common difference.

nth term: uₙ = u₁ + (n − 1)d
sum: Sₙ = (n/2)(2u₁ + (n − 1)d), or Sₙ = (n/2)(u₁ + uₙ) when you already know the last term

That (n − 1), not n, is deliberate: reaching the 5th term takes only 4 steps. Writing u₁ + nd is such a common slip that examiners expect it.

The second sum formula is the better one whenever the final term is given or easy to find — it avoids a whole line of algebra.

Geometric — a constant ratio

Each term is the previous one multiplied by a fixed number r, the common ratio.

nth term: uₙ = u₁ r^(n−1)
sum: Sₙ = u₁(r^n − 1)/(r − 1), equivalently u₁(1 − r^n)/(1 − r)

The two sum forms are identical — multiply top and bottom by −1. Use whichever keeps you out of negatives: the first when r > 1, the second when r < 1.

Finding r from two non-adjacent terms. If you are given u₃ and u₇, then u₇/u₃ = r⁴, so r = ⁴√(u₇/u₃). Count the gaps, not the terms — from the 3rd to the 7th is four steps. And note that an even root admits a negative solution: if r⁴ = 16, then r = ±2, and you must check both against the question before discarding one.

The infinite geometric sum

If — and only if — |r| < 1, the terms shrink toward zero fast enough that the total settles on a finite value:

S∞ = u₁/(1 − r)

The condition is worth a mark on its own. A question that asks you to find S∞ is very often also asking, in an earlier or later part, for the values of x for which the sum exists. That is |r| < 1, solved as an inequality — and the answer is an interval, not a single number.

If |r| ≥ 1 the sum does not exist. Say "does not converge", not "equals infinity".

Sigma notation

Σ from r = 1 to n of uᵣ just means "add these up". The letter underneath is a counter with no meaning outside the sum. The only real skill is reading off the first term and the number of terms: a sum from r = 4 to r = 20 has 17 terms, not 16. Count inclusively.

2. Exponents and logarithms

A logarithm is an exponent. log_a b = x and a^x = b say exactly the same thing, and being able to flip between them on sight is most of what this section asks.

Laws of exponents: a^m × a^n = a^(m+n), a^m ÷ a^n = a^(m−n), (a^m)^n = a^(mn), a^0 = 1, a^(−n) = 1/a^n, a^(m/n) = ⁿ√(a^m).

Laws of logarithms: log(xy) = log x + log y, log(x/y) = log x − log y, log(x^n) = n log x, and change of base log_a x = (log_b x)/(log_b a).

Two traps, both routinely examined:

log(x + y) does not simplify. There is no law for the log of a sum. Most wrong answers in this section begin by inventing one.
Solving a log equation can produce values that make the original expression undefined. log of a negative or of zero does not exist, so every solution must be substituted back and checked. Discarding an invalid root is often worth its own mark, and you must show you rejected it.

To solve an equation with the unknown in the exponent, take logs of both sides and bring the power down with log(x^n) = n log x. That third law is the workhorse of the whole section.

3. Proof

Standard level asks for simple deductive proof; higher level adds induction, contradiction and counterexample.

The habit that earns the marks: start from one side and transform it until it becomes the other. Do not start from the thing you are proving and work down to 0 = 0 — that assumes what you were asked to establish, and schemes penalise it.

Write in lines that follow from one another, and close with a sentence saying what you have shown. The reasoning marks are for the words, not the algebra.

To disprove a general statement you need exactly one counterexample, stated explicitly and shown to fail. One is enough. Conversely, no number of confirming examples ever proves a general claim — a question that says prove is never satisfied by checking n = 1, 2, 3.

4. The binomial theorem

(a + b)^n = Σ from r = 0 to n of C(n,r) a^(n−r) b^r, where C(n,r) = n!/(r!(n−r)!) is the binomial coefficient, written nCr on a calculator.

Almost every binomial question is really the same question: find one particular term, not the whole expansion. So work with the general term

C(n,r) a^(n−r) b^r

and find the r you need, rather than expanding everything and hunting.

Three things to watch:

r counts from 0. The term in b³ has r = 3, but it is the fourth term. If a question asks for "the fourth term", set r = 3.
Brackets. In (2x − 3)⁵, the a is 2x and the b is −3. Both the coefficient and the sign must be raised to the power. Forgetting to raise the 2 in 2x is the standard error.
"Constant term" means the power of x is zero. Set the total exponent of x to 0, solve for r, then evaluate that one term.

Higher level extensions

Higher level adds, in outline:

Counting principles — permutations and combinations, and the extension of the binomial theorem to fractional and negative indices, which requires |x| < 1 to converge.
Partial fractions — splitting a rational expression into simpler pieces, used again in Topic 5 for integration.
Complex numbers — Cartesian form a + bi, then modulus–argument and Euler form re^(iθ). Multiplication is where polar form pays: multiply the moduli, add the arguments.
De Moivre's theorem and complex roots, which sit evenly spaced on a circle.
Proof by induction — the four-part ritual: base case, assumption, inductive step, conclusion. Each part carries marks, including the concluding sentence, which students routinely omit.
Systems of linear equations, including when a system has no solution or infinitely many.

What actually loses marks in this topic

Using uₙ where Sₙ was asked, or the reverse.
u₁ + nd instead of u₁ + (n − 1)d.
Miscounting terms in a sigma sum — the endpoints are inclusive.
Giving S∞ without checking, or without stating, that |r| < 1.
Inventing a law for log(x + y).
Not rejecting log solutions that make the argument negative or zero.
Losing the coefficient inside a bracket in a binomial expansion.
Rounding on Paper 1. If it says exact, leave the surd, the fraction or the logarithm alone.

Educerie · written from the published IB syllabus structure for Mathematics: analysis and approaches Topic 1, first assessment 2021. Original text; all formulae quoted are standard results that appear in the IB formula booklet. Sources consulted: none beyond the published topic structure. Last reviewed 5 September 2026.