P1 6 Series | A Level Maths 9709

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🎯What you need to be able to do

Expand \( (a + b)^{n} \) for positive integer \(n\) using the binomial theorem.
Find a particular term or coefficient without writing out the whole expansion.
Use \( \binom{n}{r} \) and Pascal’s triangle, and know when each is quicker.
Recognise an arithmetic progression and use the \(n\)th term and sum formulae.
Recognise a geometric progression and use the \(n\)th term and sum formulae.
Decide whether a geometric series converges and find the sum to infinity.
Solve problems where two terms are given, or where a sum is given and \(n\) is unknown.

📚The mathematics

The binomial theorem

\[ (a + b)^{n} = a^{n} + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^{2} + \cdots + b^{n}, \qquad \binom{n}{r} = \frac{n!}{r!\,(n-r)!} \]

The powers of \(a\) count down while the powers of \(b\) count up, and in every term they add to \(n\). That last fact is the check that catches most expansion errors.

For small \(n\), Pascal’s triangle gives the coefficients faster than the formula. For \( n = 5 \) they are 1, 5, 10, 10, 5, 1. Beyond about \( n = 7 \) the formula or the calculator’s \( \binom{n}{r} \) key is quicker.

Finding one term. Most questions do not want the whole expansion. Write the general term

\[ \binom{n}{r}\,a^{\,n-r}\,b^{\,r} \]

substitute the actual \(a\) and \(b\), collect the powers of the variable, set that power equal to the one you want, and solve for \(r\). Then evaluate that single term.

The whole of each part goes into the formula, signs and coefficients included. In \( \left(2x - \dfrac{3}{x}\right)^{6} \), \( a = 2x \) and \( b = -\dfrac{3}{x} \) — not \( x \) and \( \dfrac{3}{x} \). The 2 gets raised to a power, the minus sign produces the alternating signs, and the \( \dfrac{1}{x} \) reduces the power of \(x\). And “the coefficient of \( x^{3} \)” means the entire numerical factor, not just the \( \binom{n}{r} \).

Arithmetic progressions

A fixed common difference \(d\) is added each time.

\( u_n = a + (n-1)d \)

\( S_n = \tfrac{n}{2}\left[2a + (n-1)d\right] \)

\( S_n = \tfrac{n}{2}(a + l) \)

where \(a\) is the first term and \(l\) the last. Use the third form whenever the last term is known — it is faster and there is less to get wrong.

The \( (n-1) \) is the standard slip: the tenth term involves nine steps from the first. When two terms are given, subtract the equations rather than guessing: from \( u_5 = 17 \) and \( u_{12} = 45 \) you get \( 7d = 28 \) immediately.

Geometric progressions

A fixed common ratio \(r\) multiplies each time.

\( u_n = ar^{\,n-1} \)

\( S_n = \dfrac{a\left(1 - r^{n}\right)}{1-r} \)

\( S_n = \dfrac{a\left(r^{n} - 1\right)}{r-1} \)

The last two are the same formula; take the one that keeps the denominator positive, since a negative denominator invites a sign error. To find \(r\) from two given terms, divide rather than subtract: \( \dfrac{u_7}{u_4} = r^{3} \). Dividing is what makes geometric problems short.

Beware the sign of \(r\). If \( r^{2} = 9 \) then \(r\) could be 3 or \( -3 \), and both may be valid unless the question excludes one. If \( r^{3} = 8 \) there is only the real cube root \( r = 2 \).

Sum to infinity

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

The condition \( |r| < 1 \) is the convergence condition and stating it is frequently a mark in itself. If \( |r| \ge 1 \) the sum does not exist, and producing a number anyway is worse than writing nothing. Note that negative ratios are allowed: \( r = -0.6 \) converges perfectly well, with the partial sums oscillating in towards the limit.

A common question type gives you \( S_\infty \) and one other fact and asks for \(a\) and \(r\). Set up two equations and solve simultaneously — usually by substitution, and usually ending in a quadratic in \(r\).

Which progression is it?

Arithmetic if consecutive terms differ by a constant; geometric if they are in a constant ratio. In context: a salary rising by a fixed sum each year is arithmetic; rising by a fixed percentage is geometric. Choosing wrongly is the single most consequential decision on this topic, and checking two consecutive gaps against two consecutive ratios settles it in seconds.

✏️Worked example

(a) Find the coefficient of \( x^{2} \) in the expansion of \( \left(3x - \dfrac{2}{x}\right)^{6} \). (b) The third term of a geometric progression is 20 and the sixth term is 2.5. Find the common ratio, the first term, and the sum to infinity, justifying that it exists.

(a) Here \( a = 3x \), \( b = -\dfrac{2}{x} \) and \( n = 6 \). The general term is

\[ \binom{6}{r}(3x)^{6-r}\left(-\frac{2}{x}\right)^{r} = \binom{6}{r}\,3^{6-r}\,(-2)^{r}\,x^{6-r}\,x^{-r} \]

so the power of \(x\) is \( 6 - r - r = 6 - 2r \). Setting \( 6 - 2r = 2 \) gives \( r = 2 \). That term is

\[ \binom{6}{2}\,3^{4}\,(-2)^{2}\,x^{2} = 15 \times 81 \times 4 \times x^{2} = 4860x^{2} \]

so the coefficient is \( 4860 \).

(b) Divide the two given terms:

\[ \frac{u_6}{u_3} = \frac{ar^{5}}{ar^{2}} = r^{3} = \frac{2.5}{20} = 0.125 \]

so \( r = 0.5 \). Then from \( u_3 = ar^{2} \), \( 20 = a(0.25) \), giving \( a = 80 \).

Since \( |0.5| < 1 \) the series converges, so

\[ S_\infty = \frac{80}{1 - 0.5} = 160 \]

Check it. The terms are \( 80, 40, 20, 10, 5, 2.5, \dots \) — the third is 20 and the sixth is 2.5 as required. ✓ And for a series with \( r = \tfrac{1}{2} \) the sum to infinity is always exactly twice the first term, which 160 is. In part (a), the powers of \(x\) in each factor were \( 6-r \) and \( -r \); if they had added to \( 6 \) rather than \( 6 - 2r \) you would have forgotten that \( \dfrac{1}{x} \) carries a negative power.

Do not skip the justification. “Justifying that it exists” is worth a mark on its own, and the mark is for stating \( |r| < 1 \) — 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. Find the first three terms, in ascending powers of \(x\), of the expansion of \( (2 + 3x)^{5} \).

Coefficients from Pascal's triangle for \( n = 5 \): 1, 5, 10, … With \( a = 2 \), \( b = 3x \): first term \( 2^{5} = 32 \); second \( 5(2)^{4}(3x) = 5 \times 16 \times 3x = 240x \); third \( 10(2)^{3}(3x)^{2} = 10 \times 8 \times 9x^{2} = 720x^{2} \). So \( 32 + 240x + 720x^{2} + \cdots \). The \( 3^{2} \) in the third term is the part most often dropped.

2. An arithmetic progression has first term 7 and common difference 4. Find the number of terms needed for the sum to first exceed 500.

\( S_n = \tfrac{n}{2}\left[14 + 4(n-1)\right] = \tfrac{n}{2}(4n + 10) = 2n^{2} + 5n \). Solving \( 2n^{2} + 5n > 500 \): the equation \( 2n^{2} + 5n - 500 = 0 \) gives \( n = \dfrac{-5 + \sqrt{25 + 4000}}{4} = \dfrac{-5 + 63.44}{4} = 14.6 \). Since \(n\) must be a whole number and the sum must exceed 500, take \( n = 15 \). (Check: \( S_{15} = 450 + 75 = 525 > 500 \), while \( S_{14} = 392 + 70 = 462 \).)

3. The fourth term of an arithmetic progression is 15 and the ninth is 35. Find the first term and the sum of the first 20 terms.

Subtract the equations: \( u_9 - u_4 = 5d = 35 - 15 = 20 \), so \( d = 4 \). Then \( u_4 = a + 3d \) gives \( 15 = a + 12 \), so \( a = 3 \). Finally \( S_{20} = \tfrac{20}{2}\left[2(3) + 19(4)\right] = 10(6 + 76) = 820 \).

4. A geometric progression has \( a = 48 \) and \( S_\infty = 64 \). Find the common ratio and the fourth term.

From \( S_\infty = \dfrac{a}{1-r} \): \( 64 = \dfrac{48}{1-r} \), so \( 1 - r = \dfrac{48}{64} = 0.75 \) and \( r = 0.25 \). This satisfies \( |r| < 1 \), as it must for \( S_\infty \) to have been quoted. The fourth term is \( ar^{3} = 48(0.25)^{3} = 48 \times 0.015625 = 0.75 \).

5. In the expansion of \( \left(x + \dfrac{k}{x^{2}}\right)^{9} \), the coefficient of \( x^{3} \) is 672. Find \(k\).

General term: \( \binom{9}{r}x^{9-r}\left(\dfrac{k}{x^{2}}\right)^{r} = \binom{9}{r}k^{r}x^{9-r-2r} = \binom{9}{r}k^{r}x^{9-3r} \). For \( x^{3} \), set \( 9 - 3r = 3 \), so \( r = 2 \). That coefficient is \( \binom{9}{2}k^{2} = 36k^{2} \). Setting \( 36k^{2} = 672 \) gives \( k^{2} = 18.67 \), so \( k = \pm 4.32 \). (If the question specified \( k > 0 \), take the positive root only.)

6. The first three terms of a geometric progression are \( x \), \( x + 6 \) and \( 4x \). Find the possible values of \(x\) and the corresponding common ratios.

In a geometric progression consecutive ratios are equal: \( \dfrac{x+6}{x} = \dfrac{4x}{x+6} \). Cross-multiplying, \( (x+6)^{2} = 4x^{2} \), so \( x^{2} + 12x + 36 = 4x^{2} \), giving \( 3x^{2} - 12x - 36 = 0 \) and \( x^{2} - 4x - 12 = 0 \). Factorising, \( (x-6)(x+2) = 0 \), so \( x = 6 \) or \( x = -2 \). For \( x = 6 \): terms 6, 12, 24, ratio 2. For \( x = -2 \): terms \( -2, 4, -8 \), ratio \( -2 \). Both are genuine geometric progressions, so give both — a negative ratio is perfectly legitimate.

🔗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.

Desmos — plot the partial sums of a geometric series and watch convergence happen
Khan Academy — the binomial theorem and arithmetic/geometric series
Numberphile — on infinite series and why \( |r| < 1 \) matters