Year 12

AQA A level Mathematics [Pure 1] Series Notes

Series

— Pure Mathematics 1 · AS & A Level (9709) · Chapter 6 —
★ Exam Tip: Series is a guaranteed Paper 1 topic — binomial expansion (coefficients, "term independent of x") + AP/GP (find a, d, r, Sn, S∞). Usually one 5–8 mark question, and AP/GP also turns up in cross-topic questions (trig, sectors).
Chapter map: 6.1 Binomial expansion (Pascal) → 6.2 Binomial coefficients (ⁿCᵣ) → 6.3 Arithmetic progressions → 6.4 Geometric progressions → 6.5 Infinite GP → 6.6 AP + GP mixed

1. Binomial Expansion of (a + b)ⁿ

Binomial = “two terms”, e.g. x + 3, 5x − 2y. Expanding (a + b)ⁿ with n a positive integer = binomial expansion.

Build up: (a+b)² = a² + 2ab + b² → multiply by (a+b):

(a+b)³ = (a+b)(a² + 2ab + b²) = a³ + 2a²b + ab² + a²b + 2ab² + b³ = a³ + 3a²b + 3ab² + b³

Similarly (a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.

Pattern of powers

  • 1st term is aⁿ; then power of a goes down by 1, power of b goes up by 1 each term.
  • Every term has total index = n (e.g. n=4: a⁴, a³b, a²b², ab³, b⁴).
  • Coefficients form Pascal's triangle.
Tip: each row starts & ends with 1. Each number = sum of the two numbers above it.
n=01 n=111 n=2121 n=31331 n=414641 n=515101051 3 + 3 = 6 ⇒ 1+3 = 4, 3+1 = 4 …
Pascal's triangle — add neighbours to get the next row

So with n = 5:

(a+b)⁵ = 1a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + 1b⁵
Explore 6.1Patterns in Pascal's triangle
  • Row totals: 1, 2, 4, 8, 16, 32 … = 2ⁿ. Why? put a = b = 1 in (a+b)ⁿ ⇒ sum of coefficients = 2ⁿ.
  • Fibonacci (1,1,2,3,5,8,13…): add the numbers along the shallow diagonals of the triangle.
  • Tetrahedral numbers 1, 4, 10, 20 … (stacked-ball pyramids) appear down the 4th diagonal.
  • Other patterns: triangular numbers 1,3,6,10 on the 3rd diagonal; natural numbers 1,2,3,4… on the 2nd.
Did you know? Named after French mathematician Blaise Pascal (1623–1662). Also used in probability (Prob & Stats 1, Ch 7); in Pure 2&3 you'll expand (1+x)ⁿ for any real n.

Worked Example 6.1 — use Pascal's triangle

(a) (3x + 2)³   row n=3: 1, 3, 3, 1
1(3x)³ + 3(3x)²(2) + 3(3x)(2)² + 1(2)³ = 27x³ + 54x² + 36x + 8
(b) (5 − 2x)⁴   row n=4: 1, 4, 6, 4, 1
1(5)⁴ + 4(5)³(−2x) + 6(5)²(−2x)² + 4(5)(−2x)³ + 1(−2x)⁴ = 625 − 1000x + 600x² − 160x³ + 16x⁴
Common mistake: put the whole bracket term in brackets — (−2x)² = +4x², (−2x)³ = −8x³. Signs alternate when the 2nd term is negative.

Worked Example 6.2 — expansion times a bracket

(a) expand (1 − 2x)⁵   row n=5: 1,5,10,10,5,1
1(1)⁵ + 5(1)⁴(−2x) + 10(1)³(−2x)² + 10(1)²(−2x)³ + 5(1)(−2x)⁴ + 1(−2x)⁵ = 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵
(b) coefficient of x³ in (3 + 5x)(1 − 2x)⁵

x³ arises from: 3 × (−80x³)  and  5x × (40x²)

3 × (−80x³) = −240x³ 5x × 40x² = 200x³ coefficient = −240 + 200 = −40
Trick: for (bracket)(expansion) only pair the terms whose powers add to the one you want — don't expand everything.

2. Binomial Coefficients

Pascal's triangle is too slow when n is large ⇒ need a direct way to get coefficients. The coefficients in (1+x)ⁿ are the binomial coefficients.

Explore 6.2Calculator ⁿCᵣ

On a calculator: ⁿCᵣ button (some show (n over r)). To find C(5,2): key 5 nCr 2 = 10.

C(5,0), C(5,1) … C(5,5) = 1, 5, 10, 10, 5, 1 = row 5 of Pascal's triangle!

  • coeff of x² in (1+x)⁵ is C(5,2)
  • coeff of xʳ in (1+x)ⁿ is C(n,r)
  • coeff of the 4th term in (1+x)⁵ is C(5,3)
  • coeff of the (r+1)th term in (1+x)ⁿ is C(n,r)
Careful: the (r+1)th term has xʳ — the 1st term is x⁰ (r = 0). So “4th term” ⇒ r = 3.
Binomial theorem (Key Point 6.1)(1 + x)ⁿ = C(n,0) + C(n,1)x + C(n,2)x² + … + C(n,n)xⁿ

For (a + b)ⁿ write (a+b)ⁿ = aⁿ(1 + ba)ⁿ  (a ≠ 0), giving:

Key Point 6.2(a+b)ⁿ = C(n,0)aⁿ + C(n,1)aⁿ⁻¹b¹ + C(n,2)aⁿ⁻²b² + … + C(n,n)bⁿ
(r+1)th term = C(n,r) aⁿ⁻ʳ bʳ

Worked Example 6.3 — first four terms, ascending powers

(a) (1 + x)¹⁵
1 + C(15,1)x + C(15,2)x² + C(15,3)x³ + … = 1 + 15x + 105x² + 455x³ + …
(b) (2 − 3x)¹⁰
C(10,0)2¹⁰ + C(10,1)2⁹(−3x) + C(10,2)2⁸(−3x)² + C(10,3)2⁷(−3x)³ = 1024 − 15360x + 103680x² − 414720x³ + …

e.g. x² term: 45 × 256 × 9 = 103 680

Working out C(n,r) without a calculator

Key Point 6.3C(n,0) = 1  and  C(n,n) = 1
Memory trick: numerator starts at n and has r factors going down; denominator is r! — also r factors.
Key Point 6.4 (r < n)C(n,r) = n × (n−1) × (n−2) × … × (n−r+1)r × (r−1) × … × 3 × 2 × 1

e.g. C(5,1) = 5  |  C(5,2) = 5×42×1 = 10  |  C(5,3) = 5×4×33×2×1 = 10  |  C(5,4) = 5×4×3×24×3×2×1 = 5

Symmetry: C(n,r) = C(n,n−r) (Pascal's rows are symmetric). So C(15,12) = C(15,3) — always use the smaller r!

Factorials — alternative formula

Key Point 6.5n! = n × (n−1) × (n−2) × … × 3 × 2 × 1   (e.g. 6! = 6×5×4×3×2×1)
Key Point 6.6C(n,r) = n!r! (n − r)!
WE 6.4calculating C(n,r)

(a) C(8,4) without calculator   (b) C(n,4) in terms of n

(a) 8×7×6×54×3×2×1 = 70
(b) n(n−1)(n−2)(n−3)4×3×2×1 n(n−1)(n−2)(n−3)24
WE 6.8using the factorial formula
C(8,4) 8!4!(8−4)! = 8!4! 4! = 70
C(9,3) 9!3!(9−3)! = 9!3! 6! = 84

Finding unknowns from a coefficient

WE 6.5find n

(1 − x3)ⁿ in ascending powers of x: coefficient of x² is 4. Find positive integer n.

x² term = C(n,2)(−x/3)² = n(n−1)2 × x²9 = n(n−1)18x² n(n−1)18 = 4 n(n−1) = 72 ⇒ n² − n − 72 = 0 (n − 9)(n + 8) = 0 n = 9 or −8 ⇒ n = 9 (positive integer)
WE 6.6find k

(2 + kx)⁸: coefficient of x⁵ is 2 × coefficient of x⁴; k > 0. Find k.

x⁵: C(8,5)(2)³(kx)⁵ = 448k⁵x⁵ x⁴: C(8,4)(2)⁴(kx)⁴ = 1120k⁴x⁴ 448k⁵ = 2 × 1120k⁴ 448k⁵ − 2240k⁴ = 0 448k⁴(k − 5) = 0 k = 0 or 5 ⇒ k = 5

k = 0 rejected (given k > 0). Don't divide by k⁴ blindly — factorise.

WE 6.7expansion × bracket + estimation

(a) first 3 terms of (2 − x)(1 + 2x)⁹   (b) estimate 1.99 × 1.02⁹

(1+2x)⁹ = 1 + C(9,1)(2x) + C(9,2)(2x)² + … = 1 + 18x + 144x² + … (2 − x)(1 + 18x + 144x²) = 2(1+18x+144x²) − x(1+18x+144x²) = 2 + (36 − 1)x + (288 − 18)x² + … = 2 + 35x + 270x² + …

(b) 2 − x = 1.99 ⇒ x = 0.01 and 1 + 2x = 1.02 ✓ — so let x = 0.01:

1.99 × 1.02⁹ ≈ 2 + 35(0.01) + 270(0.01)² ≈ 2.377
WE 6.9term independent of x

Find the term independent of x in (x + 5x²)⁹.

Independent of x ⇒ the powers of x must cancel. A term is C(9,r) x⁹⁻ʳ (5/x²)ʳ = C(9,r) 5ʳ x⁹⁻³ʳ.

9 − 3r = 0 ⇒ r = 3 C(9,3) x⁶ (5/x²)³ = 84 × x⁶ × 125x⁶ = 10 500

Textbook reasoning: x⁶ and (1/x²)³ → powers 6 and 3 with the sum of the powers 9; coefficient C(9,3).

★ Exam Tip: “Term independent of x” = set total power of x to 0. Write the general term, simplify powers, solve for r, then substitute. Remember to include the number in front and its sign.

3. Arithmetic Progressions (AP)

Definition: a sequence where each term differs from the one before by a constant = the common difference d (linear sequence).

e.g. 5, 8, 11, 14, 17 …   d may be zero or negative: 10, 6, 2, −2 … and 5, 5, 5 … are APs.

Notation: a = first term  |  d = common difference  |  l = last term  |  n = number of terms
term 1 term 2 term 3 term 4 term 5
a a + d a + 2d a + 3d a + 4d
Key Point 6.7nth term = a + (n − 1)d
Remember: the multiplier of d is always one less than the term number (10th term = a + 9d).
WE 6.10number of terms

How many terms in −3, 1, 5, 9, 13, …, 237?

a = −3, d = 4, nth term = 237 237 = −3 + 4(n − 1) n − 1 = 60 n = 61
WE 6.11simultaneous equations for a and d

4th term = 7, 10th term = 16. Find a and d.

a + 3d = 7  ……(1) a + 9d = 16 ……(2) (2) − (1): 6d = 9 d = 1.5 a + 4.5 = 7 ⇒ a = 2.5, d = 1.5
WE 6.12given the nth term formula

nth term = 5 − 6n. Find a and d.

n=1: 5 − 6(1) = −1 ⇒ a = −1 n=2: 5 − 6(2) = −7 d = 2nd − 1st = −7 − (−1) = −6

The coefficient of n is d (−6) — a quick check.

Sum of an AP — series

Adding terms of a sequence gives a series.

Explore 6.3Gauss's trick

1 + 2 + 3 + … + 100 = ?  Pair first with last: 1+100 = 101, 2+99 = 101, 3+98 = 101 …

50 pairs × 101 = 5050.  (Same idea: 2+4+…+500 = 125 × 502 = 62 750; 1+2+…+n = n(n+1)2.)

123 · · · 9899100 each pair= 101
50 pairs, each 101 ⇒ 50 × 101 = 5050
Key Point 6.8 — sum of n termsSn = n2(a + l)   or   Sn = n2[2a + (n − 1)d]
Which one? know the last term l → first formula. Know d → second formula (it comes from l = a + (n−1)d).

Proof

Sn = a + (a+d) + (a+2d) + … + (l−2d) + (l−d) + l reversed: Sn = l + (l−d) + (l−2d) + … + (a+2d) + (a+d) + a add: 2Sn = (a+l) + (a+l) + … n times 2Sn = n(a + l) ⇒ Sn = n2(a + l)

Put l = a + (n−1)d ⇒ Sn = n2[2a + (n−1)d].

Key Point 6.9 — true for ALL sequencesnth term = Sn − Sn−1
WE 6.13sum of whole AP

1st term −12, 17th term 12, last term 45. Find the sum of all terms.

Step 1: d. Step 2: n (using last term). Step 3: Sn.

① d 12 = −12 + 16dd = 3/2
② n 45 = −12 + (3/2)(n−1) n − 1 = 38n = 39
③ sum S39 = 392(−12 + 45) = 643½
WE 6.14given a term and a sum

10th term = 14, S₇ = 42. Find a and d.

14 = a + 9d  ……(1) 42 = 72(2a + 6d) 6 = a + 3d  ……(2) (1) − (2): 6d = 8 ⇒ d = 4/3 sub in (1): a = 2, d = 4/3
WE 6.15given Sn formula — find a, d, nth term

Sn = 4n² + n.  (a) a and d  (b) nth term.

S₁ = 4(1)² + 1 = 5 ⇒ a = 5 S₂ = 4(2)² + 2 = 18 2nd term = 18 − 5 = 13 d = 13 − 5 = 8
Method 1 a + (n−1)d = 5 + 8(n−1)= 8n − 3
Method 2 Sn − Sn−1 = 4n² + n − [4(n−1)² + (n−1)] = 4n² + n − (4n² − 8n + 4 + n − 1) = 8n − 3

Since Sn is quadratic with no constant term for an AP: S₁ = a, S₂ − S₁ = 2nd term.

✓ Quick Recap — AP
  • nth term = a + (n−1)d  |  l = a + (n−1)d
  • Sn = ½n(a + l) = ½n[2a + (n−1)d]
  • nth term = Sn − Sn−1  (any sequence)
  • Two facts about terms ⇒ two equations in a, d ⇒ subtract

4. Geometric Progressions (GP)

Definition: each term = previous term × a constant, the common ratio r. e.g. 2, 6, 18, 54 … (r = 3).

Notation: a = first term  |  r = common ratio
term 1 term 2 term 3 term 4 term 5
a ar ar² ar³ ar⁴
Key Point 6.10nth term = a rⁿ⁻¹
Compare: AP → add d (n−1) times  |  GP → multiply by r (n−1) times. r = (term) ÷ (previous term).
WE 6.16GP — find a, 8th term, nth term

5th term = 1, r = ½. Find the 8th term and the nth term.

1 = a(½)⁴ ⇒ a = 16 8th term = 16(½)⁷ = 1/8 nth term = a rⁿ⁻¹ = 16(½)ⁿ⁻¹
WE 6.17divide the equations

2nd term = 12, 5th term = 40.5. Find a, r, and the nth term.

12 = ar  ……(1) 40.5 = ar⁴ ……(2) (2) ÷ (1): ar⁴ar = 40.512 r³ = 27/8 r = 3/2 sub in (1): a = 8 nth term = 8(3/2)ⁿ⁻¹
Trick: for GP, divide the two equations (a cancels). For AP, subtract. If the power difference is even (e.g. r²) remember ± roots!
WE 6.18given nth term

nth term = 9(−⅔)ⁿ. Find a and r.

1st = 9(−⅔)¹ = −6 2nd = 9(−⅔)² = 4 r = 2nd1st = 4−6 = −⅔ a = −6, r = −2/3

Directly from the formula: the base of the power (−⅔) is r. But a ≠ 9 — first term is found by n = 1!

Sum of a GP

Explore 6.4no calculator

S₁₀ for a = 1, r = 3:  S₁₀ = 1 + 3 + 3² + … + 3⁹

3S₁₀ = 3 + 3² + 3³ + … + 3⁹ + 3¹⁰ 3S₁₀ − S₁₀ = 3¹⁰ − 1 S₁₀ = 3¹⁰ − 12

Multiply by r ⇒ almost all terms cancel when you subtract! Same method gives S = a(1 − rⁿ)/(1 − r).

Key Point 6.11Sn = a(1 − rⁿ)1 − r   or   Sn = a(rⁿ − 1)r − 1
Tip: r = 1 is not allowed in these formulae — division by 0. (If r = 1, every term = a, so Sn = na.)
  • Use the first formula when −1 < r < 1 (easier, since rⁿ is then small)
  • Use the second when r > 1 or r ≤ −1

Proof

Sn = a + ar + ar² + … + arⁿ⁻³ + arⁿ⁻² + arⁿ⁻¹  ……(1) r × (1): rSn = ar + ar² + … + arⁿ⁻² + arⁿ⁻¹ + arⁿ ……(2) (2) − (1): rSn − Sn = arⁿ − a (r − 1)Sn = a(rⁿ − 1) Sn = a(rⁿ − 1)r − 1

Multiply top and bottom by −1 ⇒ the alternative form a(1 − rⁿ)/(1 − r).

WE 6.19sum of first 12 terms

3 + 6 + 12 + 24 + …  (a = 3, r = 2, n = 12)

S₁₂ = 3(2¹² − 1)2 − 1 = 12 285
WE 6.20ratio of sums

3rd term = 9 × first term. S₆ = k × S₂. Find k.

ar² = 9a r² = 9 ⇒ r = ±3  (a ≠ 0) a(r⁶ − 1)r − 1 = k × a(r² − 1)r − 1 k = r⁶ − 1r² − 1 r = 3 or −3 ⇒ k = 91 both times

Shortcut: k = (r⁶−1)/(r²−1) = 1 + r² + r⁴ = 1 + 9 + 81 = 91 (only r² appears, so ± doesn't matter).

✓ Quick Recap — GP
  • nth term = a rⁿ⁻¹  |  r = term ÷ previous term
  • Sn = a(1−rⁿ)/(1−r) = a(rⁿ−1)/(r−1)  (r ≠ 1)
  • Two terms given ⇒ divide to find r (use ± if even power)

5. Infinite Geometric Series

Infinite sequence: terms go on forever. Take a = 2, r = ½: 2, 1, ½, ¼, ⅛ …

Partial sums: S₁ = 2, S₂ = 3, S₃ = 3½, S₄ = 3¾, S₅ = 3⅞ … → getting closer and closer to 4.

2 1 ½ ¼ ⅛ 22
2×2 square (area 4) filled by 2 + 1 + ½ + ¼ + ⅛ + …
4Sn S∞ = 4 S₁=2
partial sums S₁…S₅ climb towards the line 4 — never reach it, but get as close as we like

So we say the sum to infinity is 4: 2 + 1 + ½ + ¼ + … = 4; the series converges to 4 (a convergent series).

Idea: “=4” isn't saying we add infinitely many numbers — it means Sn can get as close to 4 as we like by taking n large. This definition gives answers that work consistently.
Recurring decimals are infinite GPs!

0.3333… = 310 + 3100 + 31000 + … ;  S₁ = 0.3, S₂ = 0.33, S₃ = 0.333 … → ⅓  so ⅓ = 0.333…

Did you know? Simon Stevin introduced infinite decimals in 1585 (De Thiende, "The tenth"). Georg Cantor (1845–1918) showed that ℕ and the rationals are the same size of infinity, but the reals are bigger.
Explore 6.5which series converge?
a, r S∞ ?
a = 3/5, r = −2 terms grow (|r| > 1) ⇒ diverges
a = 3, r = −1/5 converges: 3 ÷ (1 + 1/5) = 5/2
a = 6, r = 2/3 converges: 6 ÷ (1/3) = 18
a = −1/2, r = −2 diverges

Condition found: converges when |r| < 1, i.e. −1 < r < 1.

Remember: as n → ∞, rⁿ → 0 only if |r| < 1. Then a(1−rⁿ)/(1−r) → a(1−0)/(1−r).
Key Point 6.12S∞ = a1 − r  provided −1 < r < 1
−101 converges ✓ r ≤ −1: divergesr ≥ 1: diverges
open circles: r = ±1 are NOT included
★ Exam Tip: Always state/check −1 < r < 1 before using S∞ = a/(1−r). If r = −1 or r = 1 it does not converge.
WE 6.21sum to infinity

First four terms: 5, 4, 3.2, 2.56. (a) r (b) S∞

r = 4/5  (2nd ÷ 1st) S∞ = 51 − 4/5 = 25
WE 6.22S₃ given, find a then S∞

r = −⅔, S₃ = 63. (a) a (b) S∞

63 = a(1 − (−⅔)³)1 − (−⅔) 63 = a × 35/275/3 a = 81 S∞ = 811 − (−⅔) = 815/3 = 48⅗
Common mistake: (−⅔)³ = −8/27, so 1 − (−⅔)³ = 1 + 8/27 = 35/27. Watch the signs when r is negative.
Recurring decimals — method: write 0.57̇ as 0.57 + 0.0057 + … (a = 0.57, r = 0.01) ⇒ S∞ = 0.57/0.99 = 19/33. Pattern: repeating block of k digits ⇒ r = 10⁻ᵏ.
Fractals: repeating shapes give GPs. Koch snowflake: perimeter grows by a factor 4/3 each step (r > 1) ⇒ → ∞, but the added areas form a GP with |r| < 1 ⇒ area → 8/5 of the original triangle. Also look up the Sierpinski triangle.

6. Further Arithmetic & Geometric Series

Explore 6.6three consecutive terms a, b, c
  • AP: same difference ⇒ b − a = c − b ⇒ 2b = a + c
  • GP: same ratio ⇒ ba = cb ⇒ b² = ac
WE 6.23AP and GP together

AP: x, y, x², …   GP: x, x², y, …   x < 0. Find (a) x, y (b) S∞ of the GP (c) S₂₀ of the AP.

AP: common diff y − x = x² − y2y = x² + x ……(1)
GP: common ratio yx² = x²xy = x³ ……(2)
(1),(2): 2x³ = x² + x 2x² − x − 1 = 0  (÷ x, as x ≠ 0) (2x + 1)(x − 1) = 0 x = −½ or 1;  x < 0 ⇒ x = −½, y = −⅛

(b) GP: a = −½, r = x = −½:

S∞ = −½1 − (−½) = −⅓

(c) AP: a = −½, d = −⅛ − (−½) = ⅜, n = 20:

S₂₀ = 202[−1 + 19(⅜)] = 61.25

Full Chapter Cheat-Sheet

✓ Binomial expansion
  • (1+x)ⁿ = C(n,0) + C(n,1)x + C(n,2)x² + … + C(n,n)xⁿ
  • (a+b)ⁿ = Σ C(n,r)aⁿ⁻ʳbʳ;  (r+1)th term = C(n,r)aⁿ⁻ʳbʳ; total power always n
  • C(n,r) = n!/[r!(n−r)!] = n(n−1)…(n−r+1)/r! ;  C(n,0) = C(n,n) = 1;  C(n,r) = C(n,n−r)
  • Pascal's triangle: row sums 2ⁿ; each entry = sum of two above
  • Term independent of x: total power of x = 0
  • Estimating e.g. 1.99 × 1.02⁹: choose x so the brackets equal the given numbers (x = 0.01), then use the first few terms
✓ Arithmetic progression
  • kth term = a + (k−1)d  |  last term l = a + (n−1)d
  • Sn = ½n(a + l) = ½n[2a + (n−1)d]
  • nth term = Sn − Sn−1  |  a, b, c in AP ⇒ 2b = a + c
✓ Geometric progression
  • kth term = a rᵏ⁻¹  |  last term = a rⁿ⁻¹  |  a, b, c in GP ⇒ b² = ac
  • Sn = a(1−rⁿ)/(1−r) = a(rⁿ−1)/(r−1)
  • Converges ⇔ −1 < r < 1;  then S∞ = a/(1−r)