HookThe 6.2% bond that vanished in five weeks
On 30 August 2023, National Savings and Investments launched a one-year fixed bond paying 6.2% — the best rate on the UK savings market in fifteen years. Savers poured in roughly £7.7 billion in five weeks before NS&I pulled it in October. Behind that headline sits a sequence: put in £10,000 and after one year you hold \(10000\times1.062\), after two years \(10000\times1.062^2\), after \(n\) years \(10000\times1.062^{\,n}\) — each term the previous one multiplied by a fixed factor. That is a geometric sequence, and if you had deposited £10,000 every year the running total would be a geometric series.
This whole chapter is the mathematics of quantities that grow or shrink by a repeated step, and the running totals of those steps. The binomial expansion (4.1) tells you what a growth factor like \((1+x)^{n}\) is without a calculator; recurrence relations and sigma notation (4.2, 4.3) describe and compress a sequence; arithmetic and geometric series (4.4, 4.5) give closed formulae for the totals; the sum to infinity answers what happens if the steps never stop; and modelling (4.6) is where you decide whether the real world actually obeys your formula — because no savings rate, and no epidemic, stays constant forever.
ModelThe binomial expansion — and when the infinite version is allowed
For a positive integer \(n\), \((a+bx)^{n}\) expands into exactly \(n+1\) terms using the binomial coefficients \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\), the numbers in Pascal's triangle: \[(a+bx)^{n}=\sum_{r=0}^{n}\binom{n}{r}a^{\,n-r}(bx)^{r}.\] The powers of \(a\) fall from \(n\) to \(0\) while the powers of \(bx\) rise from \(0\) to \(n\), and the two exponents in every term add to \(n\) — a built-in check.
The A-Level extension is that the same pattern keeps working when \(n\) is any rational number, but only as an infinite series and only within a strict range: \[(1+x)^{n}=1+nx+\dfrac{n(n-1)}{2!}x^{2}+\dfrac{n(n-1)(n-2)}{3!}x^{3}+\cdots\] valid for \(|x|<1\). For \((a+bx)^{n}\) you must first factor the leading term out, writing \((a+bx)^{n}=a^{n}\left(1+\tfrac{b}{a}x\right)^{n}\), which converges only when \(\left|\tfrac{b}{a}x\right|<1\). Quoting that validity condition is worth a mark on its own, and forgetting it is the single most common error on this topic — the series is meaningless outside its interval of convergence.
Two tasks. First expand \((2+3x)^{4}\) fully. Using \(\binom{4}{0}=1,\ \binom{4}{1}=4,\ \binom{4}{2}=6,\ \binom{4}{3}=4,\ \binom{4}{4}=1\): \[(2+3x)^{4}=2^{4}+4\cdot2^{3}(3x)+6\cdot2^{2}(3x)^{2}+4\cdot2(3x)^{3}+(3x)^{4}.\] Evaluating each: \(16+96x+216x^{2}+216x^{3}+81x^{4}\). The exponents check out — every term's powers add to 4.
Now approximate \(\sqrt{1.02}\). Use \((1+x)^{1/2}=1+\tfrac12 x+\dfrac{\tfrac12(-\tfrac12)}{2}x^{2}+\cdots=1+\tfrac12 x-\tfrac18 x^{2}+\cdots\). With \(x=0.02\) (and \(|0.02|<1\), so the series is valid): \(1+\tfrac12(0.02)-\tfrac18(0.02)^{2}=1+0.01-0.00005=1.00995\). A calculator gives \(\sqrt{1.02}=1.009950\dots\) — three correct decimal places from two terms, which is the entire point of the rational-\(n\) expansion.
MechanismSequences two ways — position formula, recurrence, and the sigma shorthand
A sequence can be pinned down in two different ways. A position-to-term (nth-term) formula gives any term directly from its position, like \(u_{n}=3n-1\); you can jump straight to \(u_{100}\). A recurrence relation instead defines each term from the one before, \(x_{n+1}=f(x_{n})\), together with a starting value — you must walk term by term, but recurrences capture processes (interest, dosing, iteration) that have no easy closed form. Sequences are then classified by behaviour: increasing if \(u_{n+1}>u_{n}\) for all \(n\), decreasing if \(u_{n+1}<u_{n}\), and periodic if the terms repeat in a fixed cycle, \(u_{n+k}=u_{n}\) for some period \(k\).
Sigma notation compresses a sum: \(\sum_{r=1}^{n}u_{r}\) means "add the terms \(u_{r}\) as \(r\) runs from 1 to \(n\)". It obeys three rules worth knowing cold: \(\sum(a_{r}+b_{r})=\sum a_{r}+\sum b_{r}\), constants pull out as \(\sum k\,a_{r}=k\sum a_{r}\), and a constant summed \(n\) times gives \(\sum_{r=1}^{n}c=nc\). The limits are inclusive, so \(\sum_{r=1}^{n}\) contains \(n\) terms, not \(n-1\) — an off-by-one that quietly wrecks final answers.
Consider the recurrence \(x_{n+1}=\dfrac{1}{1-x_{n}}\) with \(x_{1}=2\). Term by term: \(x_{2}=\dfrac{1}{1-2}=-1\); \(x_{3}=\dfrac{1}{1-(-1)}=\dfrac12\); \(x_{4}=\dfrac{1}{1-\tfrac12}=2\). We are back to the start, so the sequence is periodic with period 3, cycling \(2,\,-1,\,\tfrac12,\,2,\,-1,\,\tfrac12,\dots\).
Now evaluate \(\displaystyle\sum_{r=1}^{30}x_{r}\). One full cycle sums to \(2+(-1)+\tfrac12=\tfrac32\). Thirty terms is exactly ten complete cycles, so the sum is \(10\times\tfrac32=15\). Spotting the period turns an impossible-looking 30-term sum into a two-line answer — recognising periodicity is precisely the skill the sigma limit is testing.
ModelArithmetic series — a constant step, and Gauss's pairing trick
An arithmetic sequence adds a fixed common difference \(d\) at each step, so from a first term \(a\) the \(n\)th term is \[u_{n}=a+(n-1)d.\] The \((n-1)\) — not \(n\) — is the classic trap: the first term has had zero steps added, the second has had one, so the \(n\)th has had \(n-1\). The sum of the first \(n\) terms has a beautifully symmetric formula, \[S_{n}=\tfrac{n}{2}\big(2a+(n-1)d\big)=\tfrac{n}{2}(a+l),\] where \(l\) is the last term. The second form is Gauss's schoolboy trick: pair the first term with the last, the second with the second-last, and so on — every pair sums to \(a+l\), and there are \(\tfrac{n}{2}\) pairs.
Many exam questions run the formulae backwards, giving you a sum or a term and asking for \(n\); that turns into a linear or quadratic equation in \(n\), and you keep only the positive integer root.
A concert hall has 24 seats in the front row and 3 more seats in each row behind it, for 20 rows. This is arithmetic with \(a=24\), \(d=3\).
Seats in the back row: \(u_{20}=24+(20-1)\times3=24+57=81\). Total capacity: \(S_{20}=\tfrac{20}{2}\big(24+81\big)=10\times105=1050\) seats. And which row is the first to hold exactly 60 seats? Solve \(24+(n-1)\times3=60\Rightarrow (n-1)\times3=36\Rightarrow n-1=12\Rightarrow n=13\): the 13th row. Notice that finding \(u_{20}\) first made the sum a one-line calculation with the \(\tfrac{n}{2}(a+l)\) form.
ModelGeometric series — a constant ratio, and the sum that never ends
A geometric sequence multiplies by a fixed common ratio \(r\) each step, so \(u_{n}=ar^{\,n-1}\). The sum of the first \(n\) terms comes from a slick cancellation: write \(S_{n}=a+ar+\cdots+ar^{n-1}\), multiply by \(r\), subtract, and everything in the middle collapses, leaving \[S_{n}=\dfrac{a(1-r^{n})}{1-r},\quad r\neq1.\] If \(|r|<1\) the term \(r^{n}\) shrinks towards zero as \(n\) grows, so the running total settles on a finite limit — the sum to infinity \[S_{\infty}=\dfrac{a}{1-r},\qquad \text{valid only for } |r|<1.\] When \(|r|\ge1\) the terms do not die away and the series diverges — there is no sum to infinity, and writing one down anyway is an automatic lost mark.
First a finite sum: for the sequence \(3,6,12,\dots\) we have \(a=3,\ r=2\), so \[S_{10}=\dfrac{3(1-2^{10})}{1-2}=\dfrac{3(1-1024)}{-1}=3\times1023=3069.\]
Now a sum to infinity, used to convert a recurring decimal to a fraction. Write \(0.\overline{45}=0.454545\dots=0.45+0.0045+0.000045+\cdots\), a geometric series with \(a=0.45\) and \(r=0.01\). Since \(|0.01|<1\), \[S_{\infty}=\dfrac{0.45}{1-0.01}=\dfrac{0.45}{0.99}=\dfrac{45}{99}=\dfrac{5}{11}.\] A quick check: \(5\div11=0.4545\dots\) exactly. The convergence condition \(|r|<1\) is what makes the infinite sum legitimate — here \(r=0.01\) comfortably satisfies it.
CaseModelling — the drug that plateaus, and where the model breaks
Sequences and series earn their keep when a real process repeats a step. Regular savings, loan repayments, depreciation of a car, and repeated drug doses are all geometric at heart. The A-Level twist that examiners reward is honesty about the model: a geometric model assumes the ratio stays exactly constant, which the real world rarely allows. NS&I's 6.2% rate lasted five weeks; inflation quietly erodes the real return; a car's depreciation is not truly a fixed percentage every year. So a full-mark modelling answer states the prediction, then names a limitation and suggests a refinement — for instance replacing a fixed ratio with a varying one, or a discrete step-model with a continuous one.
The repeated-dose model is the cleanest medical example. If a patient takes a fixed dose and a fixed fraction survives each interval, the drug level in the body is a geometric series whose sum to infinity is the long-run steady state — the reason a course of tablets reaches a stable therapeutic level rather than climbing forever.
A patient takes 20 mg of a drug once a day. Each 24 hours the body clears 50%, so half of what is present survives to the next dose. Immediately after the \(n\)th dose the amount present is \[A_{n}=20\big(1+\tfrac12+\tfrac12^{2}+\cdots+\tfrac12^{\,n-1}\big)=20\cdot\dfrac{1-(\tfrac12)^{n}}{1-\tfrac12}=40\big(1-(\tfrac12)^{n}\big).\] So \(A_{1}=20,\ A_{2}=30,\ A_{3}=35,\ A_{4}=37.5\), climbing towards a ceiling. As \(n\to\infty\), \((\tfrac12)^{n}\to0\), giving a steady state of \[A_{\infty}=\dfrac{20}{1-\tfrac12}=40\text{ mg}.\] The drug plateaus at 40 mg — twice a single dose. The honest limitation: real clearance is a smooth exponential decay through each day, not a single 50% jump at dosing time, so this discrete model slightly overstates the pre-dose troughs. The continuous refinement is exponential decay, which you meet in the exponentials chapter.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Quote conditions, not just answers. On any rational-\(n\) binomial expansion, write the validity range \(\left|\tfrac{b}{a}x\right|<1\) explicitly — it is a standalone mark that most candidates leave on the table. On every sum to infinity, state \(|r|<1\) before using \(\dfrac{a}{1-r}\); if the ratio fails it, say so rather than forcing a number.
Show the standard formula, then the substitution, then the answer — examiners mark method first, so a line such as \(S_{n}=\tfrac{n}{2}(2a+(n-1)d)\) written before the numbers protects follow-through marks even if the arithmetic slips. When a question gives you a sum and asks for \(n\), expect a quadratic in \(n\) and reject the non-integer or negative root with a one-line reason. In modelling questions (4.6) the final evaluation mark almost always goes to naming a limitation — a rate that will not stay constant, a discrete model standing in for a continuous process — and proposing a sensible refinement, so never end a model on the number alone.