Topic 1: Number and algebra | IB Maths AA

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🎯What you need to be able to do

Work with numbers in the form $ a \times 10^{k} $, and never write calculator notation in an answer.
Use the $n$th term and sum formulae for arithmetic and geometric sequences, including sigma notation.
Apply geometric sequences to compound interest, depreciation, annuities and real value under inflation.
Use the laws of exponents and the laws of logarithms fluently in both directions, and change the base of a logarithm.
Solve exponential equations by taking logarithms.
Decide whether an infinite geometric series converges, and sum it when it does.
Set out a simple deductive proof from left-hand side to right-hand side, using $ \equiv $ correctly.
Expand binomials with the binomial theorem and find a particular term without expanding everything.
AHL Count with permutations and combinations; extend the binomial theorem to fractional and negative indices.
AHL Split a rational expression into partial fractions.
AHL Work with complex numbers in Cartesian, polar and Euler form, and apply De Moivre’s theorem.
AHL Prove results by mathematical induction and by contradiction, and disprove them by counterexample.
AHL Solve systems of up to three linear equations in three unknowns, including the degenerate cases.

📚The mathematics

1.1 Standard form

A number in standard form is written $ a \times 10^{k} $ with $ 1 \le a < 10 $ and $ k $ an integer. The only thing that regularly costs marks here is transcription: your calculator may display 5.2E30, but that is calculator notation and it is not accepted. Write $ 5.2 \times 10^{30} $.

1.2 Arithmetic sequences and series

An arithmetic sequence adds a fixed common difference $d$ each step. The two facts you need are in the formula booklet:

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

$ S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right) $

$ S_n = \dfrac{n}{2}\left(u_1 + u_n\right) $

The second and third are the same formula wearing different clothes. Use the third whenever you already know the last term — it is faster and it is where fewer sign errors happen.

The $ (n-1) $ is the single most common slip in this whole topic. The tenth term involves nine steps, not ten. If a question gives you two terms, subtract the equations rather than guessing: from $ u_7 = 25 $ and $ u_{12} = 45 $ you get $ 5d = 20 $ directly, so $ d = 4 $.

Sigma notation compresses a sum: $ \sum_{r=1}^{n} u_r $ means add the terms from $ r = 1 $ to $ r = n $. Read off the first term by substituting the lower limit, and count the number of terms as upper limit minus lower limit plus one.

1.3 Geometric sequences and series

A geometric sequence multiplies by a fixed common ratio $r$ each step.

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

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

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

Again the last two are identical; take the one that keeps your denominator positive, because a negative denominator invites a sign error. To find $r$ from two given terms, divide rather than subtract: $ \dfrac{u_6}{u_3} = r^{3} $. Dividing is what makes geometric problems short.

Watch for the case $ r < 0 $. If $ r^{2} = 4 $ then $ r $ could be $ 2 $ or $ -2 $, and both may be valid unless the question rules one out. If $ r^{3} = 8 $ there is only the real cube root, $ r = 2 $.

1.4 Financial applications

Compound interest is a geometric sequence in disguise. With $PV$ invested at a nominal annual rate of $r\%$ compounded $k$ times per year for $n$ years,

\[ FV = PV \times \left(1 + \frac{r}{100k}\right)^{kn} \]

Half-yearly means $ k = 2 $, quarterly $ k = 4 $, monthly $ k = 12 $. Depreciation is the same formula with a negative rate: an asset losing 15% of its value each year is multiplied by $ 0.85 $ annually. To find the real value of an investment, work out the growth and then discount it by inflation — a 6% return with 4% inflation is not a 2% real return exactly, it is $ 1.06 / 1.04 $, which is about 1.92%.

Paper 2 questions may expect you to use the built-in financial package on your calculator (the TVM solver). Learn its sign convention now: money leaving you is negative, money coming to you is positive, and getting that backwards produces a plausible-looking answer that is wrong.

1.5 & 1.7 Exponents and logarithms

The laws of exponents hold for integer and rational exponents alike:

$ a^{m}a^{n} = a^{m+n} $

$ \dfrac{a^{m}}{a^{n}} = a^{m-n} $

$ (a^{m})^{n} = a^{mn} $

$ a^{-n} = \dfrac{1}{a^{n}} $

$ a^{1/n} = \sqrt[n]{a} $

$ a^{m/n} = \sqrt[n]{a^{m}} $

A logarithm is an exponent, and that one sentence is the whole idea: $ a^{x} = b $ means exactly the same thing as $ \log_a b = x $, for $ a, b > 0 $. The natural logarithm $ \ln x $ is $ \log_{\mathrm{e}} x $. The laws of logarithms are the laws of exponents translated:

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

$ \log_a \dfrac{x}{y} = \log_a x - \log_a y $

$ \log_a x^{m} = m\log_a x $

$ \log_a x = \dfrac{\log_b x}{\log_b a} $

You need these in both directions. Going forwards, $ \log 24 = \log 8 + \log 3 $; going backwards, $ 2\ln x - \ln y $ collapses to $ \ln\!\left(\dfrac{x^{2}}{y}\right) $. Collapsing a sum of logarithms into a single logarithm is how you solve most logarithmic equations, because once you have $ \log_a A = \log_a B $ you can conclude $ A = B $.

To solve an exponential equation, take logarithms of both sides and use the power law to bring the unknown down from the exponent: from $ 3^{x} = 20 $, $ x\ln 3 = \ln 20 $, so $ x = \dfrac{\ln 20}{\ln 3} \approx 2.73 $.

Three false laws that appear every session. $ \log(x + y) $ is not $ \log x + \log y $; $ \dfrac{\log x}{\log y} $ is not $ \log\dfrac{x}{y} $; and $ (\log x)^{2} $ is not $ 2\log x $. Also check the domain at the end: logarithms are only defined for positive arguments, so a solution that forces you to take the log of a negative number must be discarded, even though the algebra produced it.

1.6 Proof

At standard level you are asked for simple deductive proof, usually laid out from the left-hand side to the right-hand side. Start with one side, transform it by steps you can justify, and arrive at the other. Never start by assuming the thing you are proving and manipulating both sides at once — that shows the statement is consistent, not that it is true.

Use $ = $ between expressions that happen to be equal for particular values, and $ \equiv $ for an identity that holds for every value of the variable. Finish with a sentence, not a stranded line of algebra: “hence LHS $ \equiv $ RHS as required”.

1.8 Infinite geometric series

An infinite geometric series converges precisely when $ |r| < 1 $, and then

\[ S_\infty = \frac{u_1}{1-r}, \qquad |r| < 1 \]

State the convergence condition explicitly — it is often a mark on its own. If $ |r| \ge 1 $ the sum does not exist, and writing down a number anyway is worse than writing nothing. Note that $ |r| < 1 $ allows negative ratios: $ r = -0.4 $ converges perfectly well, with the partial sums oscillating in towards the limit.

1.9 The binomial theorem

\[ (a+b)^{n} = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}, \qquad \binom{n}{r} = \frac{n!}{r!\,(n-r)!} \]

The powers of $a$ count down while the powers of $b$ count up, and they always sum to $n$. For small $n$ Pascal’s triangle is quicker than the formula; for anything larger, use the formula or your calculator’s $ \binom{n}{r} $ key.

Most questions do not want the whole expansion. They want one term, and the skill is choosing $r$. Write the general term $ \binom{n}{r} a^{n-r} b^{r} $, collect the powers of the variable, set that power equal to the one you want, and solve for $r$.

The negative sign belongs inside the bracket. In $ (2x - 3)^{5} $ the second term of the binomial is $ -3 $, not $ 3 $. Substitute $ b = -3 $ and let the algebra handle the alternating signs. Similarly the coefficient of $ x^{3} $ means the whole numerical factor, including the powers of 2 from $ (2x) $, not just the $ \binom{n}{r} $.

AHL 1.10 Counting principles

A permutation is an arrangement, where order matters; a combination is a selection, where it does not. $ ^{n}P_r = \dfrac{n!}{(n-r)!} $ and $ ^{n}C_r = \dbinom{n}{r} = \dfrac{n!}{r!(n-r)!} $. Ask “would swapping two of the chosen items give a genuinely different outcome?” If yes, permutation; if no, combination.

The binomial theorem extends to fractional and negative indices, but only as an infinite series and only when the expansion converges. Factorise first so the bracket has the form $ (1 + x)^{n} $:

\[ (a+b)^{n} = a^{n}\left(1 + \frac{b}{a}\right)^{n}, \qquad n \in \mathbb{Q}, \quad \left|\frac{b}{a}\right| < 1 \]

AHL 1.11 Partial fractions

With two distinct linear factors in the denominator and a numerator of lower degree, write

\[ \frac{px+q}{(x-a)(x-b)} \equiv \frac{A}{x-a} + \frac{B}{x-b} \]

multiply through by the denominator, and substitute the values that kill each bracket in turn. This is not an end in itself — it exists so you can integrate rational functions (Topic 5) and sum certain series.

AHL 1.12–1.14 Complex numbers

Define $ \mathrm{i} $ by $ \mathrm{i}^{2} = -1 $. In Cartesian form $ z = a + b\mathrm{i} $, with real part $a$, imaginary part $b$, conjugate $ z^{*} = a - b\mathrm{i} $ and modulus $ |z| = \sqrt{a^{2}+b^{2}} $. Plotted on an Argand diagram, $a$ is the horizontal coordinate and $b$ the vertical.

To divide, multiply top and bottom by the conjugate of the denominator: it is the complex version of rationalising a surd, and $ zz^{*} = |z|^{2} $ is always real.

In polar (modulus–argument) form $ z = r(\cos\theta + \mathrm{i}\sin\theta) = r\,\mathrm{cis}\,\theta $, and in Euler form $ z = r\mathrm{e}^{\mathrm{i}\theta} $. Multiplication becomes simple in these forms: multiply the moduli and add the arguments. That is exactly what De Moivre’s theorem records:

\[ \left(r\,\mathrm{cis}\,\theta\right)^{n} = r^{n}\,\mathrm{cis}(n\theta) \]

Use it for powers, and — running it backwards with $ n = \frac{1}{k} $ — for the $k$th roots of a complex number, which are equally spaced around a circle of radius $ r^{1/k} $ at angular intervals of $ 2\pi/k $.

Finally, if a polynomial has real coefficients, its non-real roots occur in conjugate pairs. Given that $ 2 + 3\mathrm{i} $ is a root you immediately know $ 2 - 3\mathrm{i} $ is too, and their sum and product give you a real quadratic factor for free.

AHL 1.15 Induction, contradiction and counterexample

Proof by induction has a fixed skeleton, and the marks follow the skeleton.

Base case. Verify the statement for the first value of $n$, usually 1. Show both sides and state that they are equal.
Inductive hypothesis. Assume the statement is true for some $ n = k $. Write it out in full.
Inductive step. Prove it for $ n = k+1 $, and somewhere in that argument you must visibly use the assumption. That is the mark examiners look for hardest.
Conclusion. State it properly: since it is true for $ n = 1 $, and truth for $ n = k $ implies truth for $ n = k+1 $, it is true for all $ n \in \mathbb{Z}^{+} $ by induction.

Proof by contradiction assumes the negation of what you want and derives an absurdity — the classic being the irrationality of $ \sqrt{2} $. Counterexample disproves a universal claim with a single case, but you must also explain why your example breaks the statement; producing the number alone does not earn the mark.

AHL 1.16 Systems of linear equations

Up to three equations in three unknowns, solved algebraically by elimination or row reduction, or with technology. Three outcomes are possible: a unique solution; infinitely many, which you must express as a general solution in terms of a parameter; or none, in which case the system is inconsistent. Geometrically these are three planes meeting at a point, in a line, or not at all — a picture that pays off in Topic 3.

✏️Worked example

The third term of a geometric sequence is 18 and the sixth term is 486. (a) Find the common ratio and the first term. (b) Find the sum of the first ten terms. (c) A second geometric sequence has the same first term but common ratio $ \tfrac{2}{3} $. Find its sum to infinity, justifying that it exists.

(a) Divide, do not subtract. $ \dfrac{u_6}{u_3} = \dfrac{u_1r^{5}}{u_1r^{2}} = r^{3} $, so $ r^{3} = \dfrac{486}{18} = 27 $ and $ r = 3 $. There is only one real cube root, so no second case to consider here. Then from $ u_3 = u_1r^{2} $, $ 18 = 9u_1 $, giving $ u_1 = 2 $.

(b) With $ r > 1 $, use the version with the positive denominator:

\[ S_{10} = \frac{2\left(3^{10}-1\right)}{3-1} = \frac{2(59049-1)}{2} = 59048 \]

(c) The new sequence has $ u_1 = 2 $ and $ r = \tfrac{2}{3} $. Since $ \left|\tfrac{2}{3}\right| < 1 $ the series converges, so

\[ S_\infty = \frac{2}{1 - \tfrac{2}{3}} = \frac{2}{\tfrac{1}{3}} = 6 \]

Check it. The terms of the second sequence are $ 2, \tfrac{4}{3}, \tfrac{8}{9}, \dots $ The first three already total $ 4.22 $, and they are shrinking fast, so a limit of 6 is entirely believable. A sum to infinity that comes out smaller than the first few terms is a sign you have inverted the fraction.

Do not skip the justification in (c). The instruction “justifying that it exists” is worth a mark by itself, and the mark is for stating $ \left|r\right| < 1 $ — not for the arithmetic that follows. Whenever a question says justify, show that or hence, there is a mark attached to the reasoning rather than the number.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. An arithmetic sequence has $ u_4 = 17 $ and $ u_{11} = 45 $. Find $d$, $u_1$, and the sum of the first 20 terms.

Subtracting the two equations removes $u_1$: $ u_{11} - u_4 = 7d = 45 - 17 = 28 $, so $ d = 4 $. Then $ u_4 = u_1 + 3d $ gives $ 17 = u_1 + 12 $, so $ u_1 = 5 $. Finally $ S_{20} = \tfrac{20}{2}\left(2(5) + 19(4)\right) = 10(10 + 76) = 860 $.

2. Solve $ \log_2(x+3) + \log_2(x) = 2 $.

Collapse the left side: $ \log_2\!\left(x(x+3)\right) = 2 $, so $ x^{2} + 3x = 2^{2} = 4 $, giving $ x^{2} + 3x - 4 = 0 $ and $ (x+4)(x-1) = 0 $. So $ x = -4 $ or $ x = 1 $. Now check the domain: $ x = -4 $ would require $ \log_2(-4) $, which is undefined, so it must be rejected. The only solution is $ x = 1 $. Discarding the invalid root is part of the answer, not an afterthought.

3. Find the coefficient of $ x^{4} $ in the expansion of $ (3x - 2)^{6} $.

The general term is $ \binom{6}{r}(3x)^{6-r}(-2)^{r} $. The power of $x$ is $ 6 - r $, so for $ x^{4} $ take $ r = 2 $. That term is $ \binom{6}{2}(3x)^{4}(-2)^{2} = 15 \times 81x^{4} \times 4 = 4860x^{4} $. The coefficient is $ 4860 $ — note it includes the $ 3^{4} $ and the $ (-2)^{2} $, not just the $ \binom{6}{2} $.

4. A sum of $8000 is invested at a nominal annual rate of 5.4% compounded monthly. Find the value after 6 years, and the least whole number of years for the investment to exceed $12 000.

Value: $ 8000\left(1 + \tfrac{0.054}{12}\right)^{72} = 8000(1.0045)^{72} = \$11\,050 $ to the nearest dollar. For the second part solve $ 8000(1.0045)^{12n} > 12000 $, so $ (1.0045)^{12n} > 1.5 $. Taking logarithms, $ 12n\ln(1.0045) > \ln 1.5 $, giving $ n > \dfrac{\ln 1.5}{12\ln 1.0045} = 7.52 $. Since $n$ must be a whole number of years, $ n = 8 $. Rounding 7.52 down to 7 here would be wrong — the inequality demands the next whole year up.

5. AHL Prove by induction that $ \sum_{r=1}^{n} r(r+1) = \dfrac{n(n+1)(n+2)}{3} $ for all $ n \in \mathbb{Z}^{+} $.

Base case $ n = 1 $: LHS $ = 1 \times 2 = 2 $; RHS $ = \dfrac{1 \times 2 \times 3}{3} = 2 $. Equal, so true for $ n = 1 $.
Hypothesis: assume $ \sum_{r=1}^{k} r(r+1) = \dfrac{k(k+1)(k+2)}{3} $ for some $ k \in \mathbb{Z}^{+} $.
Step: $ \sum_{r=1}^{k+1} r(r+1) = \dfrac{k(k+1)(k+2)}{3} + (k+1)(k+2) $ — this is where the hypothesis is used. Factor out $ (k+1)(k+2) $: $ = (k+1)(k+2)\left[\dfrac{k}{3} + 1\right] = (k+1)(k+2)\cdot\dfrac{k+3}{3} = \dfrac{(k+1)(k+2)(k+3)}{3} $, which is the formula with $k+1$ in place of $k$.
Conclusion: true for $ n = 1 $, and truth for $ n = k $ implies truth for $ n = k+1 $, so true for all $ n \in \mathbb{Z}^{+} $ by mathematical induction.

6. AHL Let $ z = 1 + \mathrm{i}\sqrt{3} $. Write $z$ in polar form and hence find $ z^{6} $.

$ |z| = \sqrt{1 + 3} = 2 $, and $ \arg z = \arctan\!\left(\dfrac{\sqrt3}{1}\right) = \dfrac{\pi}{3} $, which is in the first quadrant as the Argand diagram confirms. So $ z = 2\,\mathrm{cis}\dfrac{\pi}{3} $. By De Moivre, $ z^{6} = 2^{6}\,\mathrm{cis}\left(6 \times \dfrac{\pi}{3}\right) = 64\,\mathrm{cis}(2\pi) = 64 $. The answer is real, which is worth noticing: raising a number with argument $ \pi/3 $ to the sixth power lands you exactly back on the positive real axis.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Khan Academy — sequences, series and logarithms
3Blue1Brown — the visual account of $ \mathrm{e}^{\mathrm{i}\pi} $ and complex exponentials
Desmos — plot partial sums of a geometric series and watch convergence happen