Series

• Pause and recall. Without looking above: how many terms does \((p-2q)^6\) have, what is the coefficient of the third term, and is that term positive or negative? — Seven terms; \(\binom62=15\); the third term is \(r=2\), so it carries \((-2q)^2\) and is positive.
• Pause and recall. In \(\left(2x-\dfrac1{x^2}\right)^{9}\), what power of \(x\) does the term with index \(r\) carry? — \((2x)^{9-r}\) gives \(x^{9-r}\) and \(\left(-x^{-2}\right)^{r}\) gives \(x^{-2r}\), so the power is \(9-3r\).
• Pause and recall. Is \(5,\,5,\,5,\,5,\ldots\) arithmetic, geometric, or both? — Both. The common difference is \(0\) and the common ratio is \(1\). It is the one sequence that satisfies both definitions, and it is exactly the case where the geometric sum formula breaks down, as section D explains.
• Pause and recall. An AP has \(u_1=12\) and \(u_5=0\). What is \(d\), and what is \(S_5\)? — Four steps take you from 12 to 0, so \(d=-3\); and \(S_5=\frac52(12+0)=30\). The last term being zero is not a problem for an AP — only for a GP.
• Pause and recall. A GP has \(u_2=6\) and \(u_4=54\). What are the possible values of \(r\)? — \(r^2=54/6=9\), so \(r=3\) or \(r=-3\). An even gap between the given terms leaves the sign of \(r\) undetermined, and both progressions are legitimate answers unless the question rules one out.
• Pause and recall. A population of 5000 grows by 4% each year. What is the population after \(n\) years, and is the increase each year an AP or a GP? — \(5000(1.04)^n\); and the yearly increases are \(200,\,208,\,216.32,\ldots\), themselves a GP with ratio \(1.04\). The increases of a GP form another GP; the increases of an AP are constant.
• Pause and recall. A GP has \(S_\infty=40\) and \(a=10\). Find \(r\). — \(\frac{10}{1-r}=40\) gives \(1-r=\frac14\), so \(r=\frac34\). Check \(\left\lvert\frac34\right\rvert<1\) ✓, which confirms the answer is self-consistent: a sum to infinity was quoted, so \(r\) had better satisfy the condition.
• Interleave, do not block. When you revise this chapter alongside others, mix binomial questions with progression questions in the same session rather than doing ten of each. The difficulty you feel is the point: deciding which tool to use is the decision a mixed paper forces on you, and blocked practice removes that decision entirely.

• 12.1 — Expand \((a+b)^n\) for a positive integer \(n\), with correct binomial coefficients, correct falling and rising powers, and correct alternating signs when the second quantity is negative.
• 12.1 — State without hesitation that an expansion of power \(n\) has \(n+1\) terms before like terms are collected, and use the symmetry of the coefficients as a check.
• 12.2 — Write down the general term \(T_{r+1}=\binom nr a^{n-r}b^{r}\) and use it to find a specified power, a specified coefficient, or the term independent of \(x\), without expanding anything else.
• 12.2 — Recognise that \(r\) must be a whole number with \(0\le r\le n\), and say plainly that a requested power does not occur when the index equation fails that test.
• 12.3 — Decide whether a sequence is arithmetic, geometric or neither by testing differences and ratios across at least three terms, and explain the structural difference rather than pointing at the numbers.
• 12.4 — Use \(u_n=a+(n-1)d\) and \(u_n=ar^{n-1}\) to find any term, and to work backwards from two given terms to \(a\) and \(d\) or \(a\) and \(r\).
• 12.4 — Use \(S_n=\frac n2\{2a+(n-1)d\}\), \(S_n=\frac n2(a+l)\) and \(S_n=\dfrac{a(1-r^n)}{1-r}\), choosing between them from what the question actually supplies, and handle the \(r=1\) case separately.
• 12.4 — Translate a worded context into the right progression, deciding from the wording whether a quantity is the term for \(n=1\) or a starting value that sits before the sequence begins.
• 12.5 — Test \(\lvert r\rvert<1\) before writing \(S_\infty\), and use \(S_\infty=\dfrac a{1-r}\) only once that test has passed.
• 12.5 — Explain why a particular geometric progression does or does not possess a finite sum to infinity, including the case of a negative ratio that alternates in sign and still converges.

Why this topic earns its place. Series is where algebraic structure meets repeated change. Compound interest, depreciation, population models, drug half-lives and repayment schedules are all geometric progressions in disguise, while the binomial expansion is the first tool you will meet that produces an entire family of coefficients from one rule. Later, in Chapter 14, the same habit of asking “what is repeating, and does it settle down?” is exactly what a limit asks.

General Binomial Term

The single formula that produces any one term of the expansion of a plus b to the power n without expanding the rest. The term with index r is n choose r multiplied by a to the power n minus r and b to the power r, and it is the term numbered r plus one. The index r must be a whole number satisfying zero is less than or equal to r is less than or equal to n; a value outside that range means the requested term does not occur in the expansion.

Sum to Infinity

The finite total that the partial sums of an infinite geometric progression approach as more and more terms are added. It exists only when the modulus of the common ratio is less than one, and it then equals the first term divided by one minus the common ratio. When the modulus of the ratio is one or greater, the terms do not shrink towards zero, the partial sums never settle, and no sum to infinity exists at all.

Sum of an Arithmetic Progression

The total of the first n terms of a sequence with a constant common difference. It equals n over two multiplied by the bracket two a plus n minus one d, where a is the first term and d the common difference; equivalently it is n over two multiplied by a plus l, where l is the last term included. The second form is derived by writing the series forwards and backwards and pairing the columns, each of which totals a plus l.

Arithmetic Progression

A sequence in which each term is obtained from the previous one by adding the same fixed number, called the common difference d. Its first term is a and its nth term is a plus n minus one times d, because the nth term has taken n minus one steps from the first. The defining test is that the difference between consecutive terms is constant across the whole sequence.

Repeated Percentage Change

A situation in which a quantity changes by the same percentage of its current value in every period, so the same multiplier is applied again and again. If the percentage is p written as a decimal, the multiplier is one plus p for a gain and one minus p for a loss, and the value after n periods is the initial amount multiplied by that multiplier raised to the power n. Because a fixed percentage of a changing amount is not a fixed amount, this always produces a geometric progression and never an arithmetic one.

Binomial Theorem

A rule that expands a bracket of the form (a + b) raised to a positive whole-number power n into a sum of n + 1 terms. The term for index r is the binomial coefficient n choose r multiplied by a to the power n minus r and b to the power r, so the power of the first quantity falls from n to 0 while the power of the second rises from 0 to n, and the two powers add to n in every term.

Sum of a Geometric Progression

The total of the first n terms of a sequence with a constant common ratio. It equals a multiplied by one minus r to the power n, all divided by one minus r, where a is the first term and r the common ratio, and it is valid only when r is not equal to one. The equivalent form a times r to the power n minus one, all over r minus one, is the same expression with the signs of numerator and denominator both reversed. When r equals one every term is a, so the sum is n times a.

Geometric Progression

A sequence in which each term is obtained from the previous one by multiplying by the same fixed number, called the common ratio r. Its first term is a and its nth term is a times r to the power n minus one. The defining test is that the ratio of consecutive terms is constant, which may only be computed where the preceding term is not zero.

• 1. Writing \(n\) terms in an expansion of power \(n\). Why it fails The index \(r\) runs \(0,1,\ldots,n\), and that list contains \(n+1\) values, not \(n\). The missing term is almost always one of the two ends. Fix Count the terms before simplifying. \((2x-3)^4\) must show five.
• 2. Ignoring the alternating signs in \((a-b)^n\). Why it fails The theorem expands a sum. Writing \((a-b)^n\) means \(b\) is really \(-b\), and \((-b)^r\) is negative for odd \(r\). Fix Rewrite the bracket as \(\bigl(a+(-b)\bigr)^n\) and carry the minus sign inside brackets all the way through.
• 3. Omitting the binomial coefficient. Why it fails \(\binom nr\) counts how many ways that combination of \(a\)s and \(b\)s can arise. Dropping it makes every middle term too small. Fix Check the symmetry: \(1,4,6,4,1\) reads the same backwards. A run without that symmetry has lost a coefficient.
• 4. Calling \(T_{r+1}\) the \(r\)th term. Why it fails \(r=0\) produces the first term, so the term number always runs one ahead of the index. Fix For the \(k\)th term use \(r=k-1\); when you find \(r\), report the term as the \((r+1)\)th.
• 5. Accepting a non-integer index, such as \(r=\frac73\). Why it fails \(r\) counts how many factors of \(b\) a term contains. There is no term containing two and a third factors, so no such term exists. Fix Say so explicitly: “\(r\) is not an integer, so there is no term in \(x^5\).” That statement is the answer.
• 6. Accepting an index outside \(0\le r\le n\). Why it fails The expansion stops at \(r=n\). An index of \(7\) in an expansion of power \(6\) refers to a term that was never generated. Fix Run both tests every time — whole number and in range. Each catches a different failure.
• 7. Expanding the whole bracket when one term was asked for. Why it fails Not an error of truth, but of time and exposure: ten unnecessary terms are ten chances to slip. Fix If the words are “the coefficient of”, “the term in” or “independent of\(\,x\)”, go straight to the general term.
• 8. Confusing the binomial \(a\) with a progression's first term \(a\). Why it fails In \((2x-3)^4\) the symbol \(a\) stands for the expression \(2x\); in a progression \(a\) is a number. Carrying a habit from one into the other produces nonsense that is hard to spot. Fix Declare your symbols in one line before you start: “let \(a=2x\), \(b=-3\), \(n=4\)”.
• 9. Treating greatest-term questions as required content. Why it fails Finding the numerically greatest term of an expansion, and separate theoretical properties of the coefficient sequence, are outside outcomes 12.1–12.2 for 4037. Fix Spend the revision time on the general term instead. The coefficient symmetry is used here only as a check.
• 10. Forgetting to raise a numerical factor to its power. Why it fails In \(\left(x^2+\frac2x\right)^6\) the \(2\) becomes \(2^r\). Writing \(\binom6r\,2\,x^{12-3r}\) gets the power of \(x\) right and the value wrong by a factor of 8. Fix Keep the whole of \(b\) inside a bracket: \(\left(\frac2x\right)^r\), then expand it.
• 11. Deciding AP or GP from only two terms. Why it fails Any two non-zero numbers fit both patterns: \(4,7\) is an AP with \(d=3\) and a GP with \(r=\frac74\). Fix Test across three terms — two differences or two ratios — before committing.
• 12. Using \(a+nd\) instead of \(a+(n-1)d\). Why it fails The first term has taken no steps, so \(u_1=a\). \(a+nd\) gives \(a+d\) at \(n=1\), which is the second term. Fix Test any \(n\)th-term formula at \(n=1\); it must return \(a\) exactly.
• 13. Using \(ar^n\) instead of \(ar^{\,n-1}\). Why it fails The same step-counting argument: \(u_1=ar^0=a\). \(ar^n\) is the term after the one you wanted. Fix Write the first four terms out: \(a,ar,ar^2,ar^3\). The exponent is visibly one behind the position.
• 14. Confusing \(u_n\) with \(S_n\). Why it fails They answer different questions and differ wildly in size: for \(a=5,d=3\), \(u_{20}=62\) but \(S_{20}=670\). Fix Write the symbol you are computing at the top of the working. If the words contain “total” or “altogether”, it is a sum.
• 15. Losing the sign of a negative common ratio. Why it fails \(\left(-\frac13\right)^5=-\frac1{243}\), so \(1-r^5=1+\frac1{243}\). Dropping the sign changes both the terms and the sum. Fix Always bracket a negative ratio before raising it to a power, and expect the terms to alternate as a check.
• 16. Dividing consecutive terms when the earlier one is zero. Why it fails \(r=u_{n+1}/u_n\) requires \(u_n\ne0\). A GP with \(a\ne0\) never contains a zero term anyway, so a zero is evidence the sequence is not geometric. Fix State the restriction when you use the ratio test, and treat a zero term as a signal to test for an AP instead.
• 17. Using the GP sum formula when \(r=1\). Why it fails The denominator \(1-r\) is zero, so the expression is \(\frac00\) and means nothing. Fix Handle it directly: every term is \(a\), so \(S_n=na\). Say why on the page.
• 18. Treating repeated percentage change as an AP. Why it fails A fixed percentage of a changing amount is not a fixed amount. Only the multiplier stays constant, which is the definition of a GP. Fix Convert the percentage to a multiplier immediately: 12% loss becomes \(\times0.88\), 5% gain becomes \(\times1.05\).
• 19. Using the wrong exponent in a timed context. Why it fails A value quoted before any change is the \(n=0\) value, giving an exponent of \(n\). A value quoted as the first of a list is \(u_1\), giving \(n-1\). Fix Ask which of the two the words describe, and write your decision down: “the purchase price is the value at \(n=0\)”.
• 20. Using \(S_\infty\) without first showing \(\lvert r\rvert<1\). Why it fails \(\frac a{1-r}\) returns a number for any \(r\ne1\), including ones for which no infinite sum exists. For \(a=5,r=2\) it returns \(-5\), a negative total of positive terms. Fix Evaluate \(\lvert r\rvert\), compare it with 1 in writing, and only then substitute.
• 21. Assuming every negative ratio diverges. Why it fails The test is on \(\lvert r\rvert\), not on the sign. \(r=-\frac13\) gives \(\lvert r\rvert=\frac13<1\), so it converges — the partial sums simply approach the limit alternately from above and below. Fix Take the modulus first. Only \(r=-1\) and \(r<-1\) fail among the negatives.
• 22. Rounding an exact answer unnecessarily. Why it fails \(\frac{243}2\) is exact; “\(121.5\) (3 s.f.)” attaches an accuracy claim to something that has no error at all, and \(\frac{244}{27}\approx9.04\) genuinely throws information away. Fix Keep fractions and surds unless a decimal is asked for. Round once, at the very end, and never mix a fraction and a decimal inside one value.

• Both 4037 papers may assess any part of the subject content, so Topic 12 can appear in either. What changes between them is the kind of answer that is expected.
• The examination supplies the binomial theorem together with \(\binom nr\), and the arithmetic and geometric progression formulae — including \(S_\infty\) with the condition \(\lvert r\rvert<1\) printed beside it, exactly as \(S_n\) is printed with \(r\ne1\). What the list cannot supply is the use of those conditions: it will not evaluate \(\lvert r\rvert\) for you, decide whether this progression is arithmetic or geometric, or read a worded context. Treat the list as a spelling aid, not as a method.
• Write down — no working expected; the answer is one substitution away. Show that — the target value is printed, so every step must be visible and you must carry more accuracy than the printed answer displays. Explain — a reason is the answer; for convergence this means quoting \(\lvert r\rvert<1\) and evaluating it, not merely asserting “it converges”. Determine — establish the result with certainty, which in a GP usually means resolving the \(r\ne1\) or \(\lvert r\rvert<1\) question first.
• Accuracy contract for this topic. Exact answers may need fractions or \(\pi\); a non-exact answer is given to at least 3 significant figures; intermediate values are kept unrounded. Never write a fraction and a decimal inside the same value, and never round \(\frac{243}{2}\) to “\(121.5\) (3 s.f.)” — it is exact, so no accuracy statement belongs on it.