Mathematics

Answer Key

Sequences

Pack A — Answers

# Question Answer
1 Write down the next two terms in the sequence: $3,\ 7,\ 11,\ 15,\ \ldots$. 19, 23
2 Write down the next two terms in the sequence: $2,\ 6,\ 18,\ 54,\ \ldots$. 162, 486
3 State whether the sequence $4,\ 8,\ 12,\ 16,\ \ldots$ is arithmetic, geometric, or neither. Arithmetic ($d = 4$)
4 For the sequence $15,\ 11,\ 7,\ 3,\ \ldots$, state the common difference $d$. $d = -4$
5 For the sequence $64,\ 32,\ 16,\ 8,\ \ldots$, state the common ratio $r$. $r = \dfrac{1}{2}$
6 A sequence begins $7,\ 10,\ 13,\ 16,\ \ldots$. Write down (a) $u_1$ and (b) $u_3$. (a) $u_1 = 7$ (b) $u_3 = 13$
7 A sequence has first term $u_1 = 5$ and term-to-term rule "add $3$". Write down the first four terms. 5, 8, 11, 14
8 A sequence has $u_1 = 2$ and term-to-term rule "multiply by $3$". Write down the first four terms. 2, 6, 18, 54
9 The $n$th term of a sequence is $u_n = 3n + 2$. Find $u_1$ and $u_5$. $u_1 = 5$, $u_5 = 17$
10 Find the 6th term of the sequence $2,\ 5,\ 8,\ 11,\ \ldots$. 17
11 Find the $n$th term formula for the arithmetic sequence $5,\ 8,\ 11,\ 14,\ \ldots$. $u_n = 3n + 2$
12 Find the $n$th term formula for the sequence $20,\ 17,\ 14,\ 11,\ \ldots$. $u_n = 23 - 3n$
13 For the arithmetic sequence with $u_1 = 7$ and $d = 4$, find $u_{20}$. $u_{20} = 83$
14 For the sequence with $n$th term $u_n = 4n + 1$, find which term equals $81$. $n = 20$ (the 20th term)
15 A geometric sequence has $u_1 = 3$ and $r = 2$. Find $u_5$. $u_5 = 48$
16 A sequence is defined recursively by $u_1 = 4$ and $u_{{n+1}} = u_n + 5$. Find the first four terms. 4, 9, 14, 19
17 A sequence is defined by $u_1 = 5$ and $u_{{n+1}} = 2 u_n$. Find $u_4$. $u_4 = 40$
18 A pattern uses $4$ matchsticks for shape 1, and each new shape adds $3$ more. How many matchsticks for shape 10? 31
19 An arithmetic sequence has $u_3 = 11$ and $u_8 = 31$. Find the common difference. $d = 4$
20 Find the $n$th term formula for the geometric sequence $4,\ 12,\ 36,\ 108,\ \ldots$. $u_n = 4 \cdot 3^{n-1}$
21 An arithmetic sequence has $u_5 = 17$ and $u_{15} = 47$ where $n = 15$. Find $u_1$ and $d$. $u_1 = 5$, $d = 3$
22 The $n$th term of a sequence is $u_n = 7n + 3$. Find the first term greater than $100$. $u_{14} = 101$
23 A sequence is defined by $u_1 = 5$ and $u_{{n+1}} = 2u_n - 3$. Find $u_4$. $u_4 = 19$
24 A geometric sequence has $u_2 = 6$ and $u_4 = 54$ (both positive). Find $u_1$ and $r$. $u_1 = 2$, $r = 3$
25 A sequence is given by $1,\ 4,\ 9,\ 16,\ 25,\ \ldots$. Find the next term and describe the pattern. 36; square numbers $u_n = n^2$
26 A sequence is defined by $u_1 = 1$, $u_2 = 2$ and $u_{{n+1}} = u_n + u_{{n-1}}$. Find $u_6$. $u_6 = 13$
27 Which term of the sequence $u_n = 8n - 5$ equals $99$? $n = 13$
28 The first three terms of a sequence are $7,\ 4,\ 1,\ -2,\ \ldots$. (a) State the type. (b) Find $u_n$. (a) Arithmetic, $d = -3$. (b) $u_n = 10 - 3n$
29 Find the sum of the first $10$ terms of the arithmetic sequence with $u_1 = 3$ and $d = 5$. $S_{10} = 255$
30 A geometric sequence has $u_1 = 16$ and $r = \dfrac{1}{2}$. Find $u_4$. $u_4 = 2$
31 An arithmetic sequence has $u_4 = 14$ and $u_{14} = 44$ where $n = 14$. Find $u_n$ as a formula. $u_n = 3n + 2$
32 Find the sum of all multiples of $7$ between 1 and $100$ inclusive. 735
33 A geometric sequence has $u_3 = 12$ and $u_6 = 96$ (both positive). Find $u_1$ and $r$. $r = 2$, $u_1 = 3$
34 Sequence A has $u_n = 5n + 2$ and sequence B has $u_n = 8n - 7$. For which value of $n$ do they have the same term? $n = 3$ (both equal 17)
35 A sequence is defined by $u_1 = 8$ and $u_{{n+1}} = \dfrac{u_n}{2} + 3$. Find $u_4$. $u_4 = 6.25$ (i.e. $\dfrac{25}{4}$)
36 Find the sum of the first $6$ terms of the geometric sequence with $u_1 = 2$ and $r = 3$. $S_6 = 728$
37 The first four terms of a sequence are $2,\ 6,\ 12,\ 20,\ \ldots$. Find the next term and a quadratic formula for $u_n$. 30; $u_n = n^2 + n$
38 A bacterial colony of $50$ cells doubles every hour. (a) Write the number of cells $u_n$ after $n$ hours. (b) After how many full hours will the population exceed $1000$? (a) $u_n = 50 \cdot 2^n$. (b) After 5 hours ($u_5 = 1600$)
39 Maya saves £$5$ in week 1 and £$3$ more each week than the previous week. In which week does her **weekly** saving first exceed £$50$? Week 17 (weekly = £53)
40 A sequence is defined by $u_1 = 1$ and $u_{{n+1}} = u_n + 2n + 1$. Find $u_5$. $u_5 = 25$

Pack B — Answers

# Question Answer
1 Write down the next two terms in the sequence: $5,\ 9,\ 13,\ 17,\ \ldots$. 21, 25
2 Write down the next two terms in the sequence: $3,\ 12,\ 48,\ 192,\ \ldots$. 768, 3072
3 State whether the sequence $1,\ 2,\ 4,\ 8,\ \ldots$ is arithmetic, geometric, or neither. Geometric ($r = 2$)
4 For the sequence $20,\ 14,\ 8,\ 2,\ \ldots$, state the common difference $d$. $d = -6$
5 For the sequence $81,\ 27,\ 9,\ 3,\ \ldots$, state the common ratio $r$. $r = \dfrac{1}{3}$
6 A sequence begins $12,\ 18,\ 24,\ 30,\ \ldots$. Write down (a) $u_1$ and (b) $u_3$. (a) $u_1 = 12$ (b) $u_3 = 24$
7 A sequence has first term $u_1 = 2$ and term-to-term rule "add $7$". Write down the first four terms. 2, 9, 16, 23
8 A sequence has $u_1 = 5$ and term-to-term rule "multiply by $2$". Write down the first four terms. 5, 10, 20, 40
9 The $n$th term of a sequence is $u_n = 4n + 1$. Find $u_1$ and $u_5$. $u_1 = 5$, $u_5 = 21$
10 Find the 6th term of the sequence $7,\ 12,\ 17,\ 22,\ \ldots$. 32
11 Find the $n$th term formula for the arithmetic sequence $3,\ 9,\ 15,\ 21,\ \ldots$. $u_n = 6n - 3$
12 Find the $n$th term formula for the sequence $50,\ 45,\ 40,\ 35,\ \ldots$. $u_n = 55 - 5n$
13 For the arithmetic sequence with $u_1 = 12$ and $d = 3$, find $u_{25}$. $u_{25} = 84$
14 For the sequence with $n$th term $u_n = 5n + 2$, find which term equals $102$. $n = 20$ (the 20th term)
15 A geometric sequence has $u_1 = 2$ and $r = 3$. Find $u_5$. $u_5 = 162$
16 A sequence is defined recursively by $u_1 = 1$ and $u_{{n+1}} = u_n + 6$. Find the first four terms. 1, 7, 13, 19
17 A sequence is defined by $u_1 = 4$ and $u_{{n+1}} = 3 u_n$. Find $u_4$. $u_4 = 108$
18 A pattern uses $5$ matchsticks for shape 1, and each new shape adds $4$ more. How many matchsticks for shape 10? 41
19 An arithmetic sequence has $u_3 = 9$ and $u_8 = 24$. Find the common difference. $d = 3$
20 Find the $n$th term formula for the geometric sequence $5,\ 10,\ 20,\ 40,\ \ldots$. $u_n = 5 \cdot 2^{n-1}$
21 An arithmetic sequence has $u_5 = 14$ and $u_{15} = 44$ where $n = 15$. Find $u_1$ and $d$. $u_1 = 2$, $d = 3$
22 The $n$th term of a sequence is $u_n = 5n + 2$. Find the first term greater than $200$. $u_{40} = 202$
23 A sequence is defined by $u_1 = 4$ and $u_{{n+1}} = 2u_n - 2$. Find $u_4$. $u_4 = 18$
24 A geometric sequence has $u_2 = 8$ and $u_4 = 32$ (both positive). Find $u_1$ and $r$. $u_1 = 4$, $r = 2$
25 A sequence is given by $1,\ 8,\ 27,\ 64,\ \ldots$. Find the next term and describe the pattern. 125; cube numbers $u_n = n^3$
26 A sequence is defined by $u_1 = 2$, $u_2 = 3$ and $u_{{n+1}} = u_n + u_{{n-1}}$. Find $u_6$. $u_6 = 21$
27 Which term of the sequence $u_n = 6n - 1$ equals $89$? $n = 15$
28 The first three terms of a sequence are $10,\ 6,\ 2,\ -2,\ \ldots$. (a) State the type. (b) Find $u_n$. (a) Arithmetic, $d = -4$. (b) $u_n = 14 - 4n$
29 Find the sum of the first $12$ terms of the arithmetic sequence with $u_1 = 2$ and $d = 3$. $S_{12} = 222$
30 A geometric sequence has $u_1 = 81$ and $r = \dfrac{1}{3}$. Find $u_4$. $u_4 = 3$
31 An arithmetic sequence has $u_4 = 11$ and $u_{14} = 41$ where $n = 14$. Find $u_n$ as a formula. $u_n = 3n - 1$
32 Find the sum of all multiples of $5$ between 1 and $200$ inclusive. 4100
33 A geometric sequence has $u_3 = 18$ and $u_6 = 486$ (both positive). Find $u_1$ and $r$. $r = 3$, $u_1 = 2$
34 Sequence A has $u_n = 5n + 2$ and sequence B has $u_n = 7n - 6$. For which value of $n$ do they have the same term? $n = 4$ (both equal 22)
35 A sequence is defined by $u_1 = 12$ and $u_{{n+1}} = \dfrac{u_n}{2} + 2$. Find $u_4$. $u_4 = 5$
36 Find the sum of the first $5$ terms of the geometric sequence with $u_1 = 3$ and $r = 2$. $S_5 = 93$
37 The first four terms of a sequence are $3,\ 8,\ 15,\ 24,\ \ldots$. Find the next term and a quadratic formula for $u_n$. 35; $u_n = n^2 + 2n$
38 A bacterial colony of $100$ cells doubles every hour. (a) Write the number of cells $u_n$ after $n$ hours. (b) After how many full hours will the population exceed $5000$? (a) $u_n = 100 \cdot 2^n$. (b) After 6 hours ($u_6 = 6400$)
39 Maya saves £$4$ in week 1 and £$2$ more each week than the previous week. In which week does her **weekly** saving first exceed £$40$? Week 20 (weekly = £42)
40 A sequence is defined by $u_1 = 2$ and $u_{{n+1}} = u_n + 2n + 1$. Find $u_5$. $u_5 = 26$

Problems — Worked Solutions

1

**Theatre seating.** A theatre has 20 seats in row 1, 23 seats in row 2, 26 seats in row 3, and so on, each row adding 3 seats. (a) How many seats are in row 12? (b) Which row is the first to have at least 50 seats? (c) How many seats are there in total in the first 12 rows?

Answer

(a) 53 (b) Row 11 (51 seats) (c) 438

Arithmetic with $u_1 = 20$, $d = 3$. So $u_n = 20 + (n-1) \times 3 = 3n + 17$. (a) $u_{12} = 3(12) + 17 = 53$ seats. (b) Solve $3n + 17 \geq 50$: $3n \geq 33$, so $n \geq 11$. Row 11 has $u_{11} = 50$ seats — first row with **at least** 50. (c) Sum: $S_{12} = \frac{12}{2}(u_1 + u_{12}) = 6(20 + 53) = 6 \times 73 = \mathbf{438}$ seats.
2

**Bouncing ball.** A ball is dropped from a height of 80 cm. Each bounce reaches $\frac{3}{4}$ of the previous bounce height. (a) Write the height of the $n$th bounce as a geometric sequence. (b) What is the height of the 5th bounce, to the nearest mm? (c) After how many bounces is the height less than 10 cm for the first time?

Answer

(a) $h_n = 80 \cdot (\tfrac{3}{4})^n$ (b) 19.0 cm (190 mm) (c) 8th bounce (8.0 cm)

(a) Geometric: drop $h_0 = 80$, $r = \frac{3}{4}$. Height of $n$th bounce: $h_n = 80 \cdot (\tfrac{3}{4})^n$. (b) $h_5 = 80 \cdot (0.75)^5 = 80 \times 0.2373 \approx 18.98$ cm ≈ **19.0 cm**. (c) Need $80 \cdot (0.75)^n < 10$, i.e. $(0.75)^n < 0.125$. Check: $(0.75)^7 \approx 0.1335$, $(0.75)^8 \approx 0.1001$. So 8th bounce: $h_8 \approx 8.0$ cm. **8th bounce**.
3

**Salary growth.** A graduate has two job offers. - **Company A:** starting salary £24,000, with annual increases of £1,500. - **Company B:** starting salary £22,000, with annual increases of 5% on the previous year's salary. (a) Write a formula for the salary $A_n$ at company A and $B_n$ at company B in year $n$. (b) In which year does B first overtake A? (c) What is the total earned at each company over the first 5 years?

Answer

(a) $A_n = 22500 + 1500n$; $B_n = 22000 \cdot (1.05)^{n-1}$ (b) Year 17 (c) A: £135,000; B: £121,550

(a) Company A is arithmetic: $A_n = 24000 + (n-1)\times 1500 = 22500 + 1500n$. Company B is geometric: $B_n = 22000 \cdot (1.05)^{n-1}$. (b) Set $B_n > A_n$. Test years 1–15: - Year 1: A = 24000, B = 22000. - Year 5: A = 30000, B = $22000 \times 1.05^4 \approx 26744$. - Year 10: A = 37500, B = $22000 \times 1.05^9 \approx 34124$. - Year 15: A = 45000, B = $22000 \times 1.05^{14} \approx 43560$. - Year 17: A = 48000, B = $\approx 48025$. Company B overtakes A in **year 17**. (c) Sum over first 5 years. - A: $S_5 = \frac{5}{2}(A_1 + A_5) = \frac{5}{2}(24000 + 30000) = 2.5 \times 54000 = \mathbf{\pounds 135{,}000}$. - B: $S_5 = \frac{22000(1.05^5 - 1)}{0.05} = \frac{22000 \times 0.2763}{0.05} \approx \mathbf{\pounds 121{,}550}$.
4

**Shifted pattern.** The diagram below shows the first three patterns made from dots. Pattern 1: 4 dots (square corners) Pattern 2: 7 dots Pattern 3: 10 dots (a) How many dots are in pattern $n$? (b) Pattern $k$ has 88 dots. Find $k$. (c) A student says "Pattern 100 has 304 dots." Is she correct? Show your reasoning.

Answer

(a) $u_n = 3n + 1$ (b) $k = 29$ (c) No — pattern 100 has 301 dots

(a) Arithmetic with $u_1 = 4$, $d = 3$. So $u_n = 4 + (n-1)\times 3 = 3n + 1$. (b) Set $3n + 1 = 88$: $3n = 87$, $n = 29$. (c) $u_{100} = 3(100) + 1 = 301$, **not 304**. The student is incorrect.
5

**Geometric shrinkage.** A piece of paper has area 800 cm². It is folded in half repeatedly so that the new exposed top area halves each time. (a) Write a sequence for the visible area $A_n$ after $n$ folds. (b) After how many folds is the visible area less than 1 cm²? (c) Could the area ever be exactly 0 cm²? Justify mathematically.

Answer

(a) $A_n = 800 \cdot (\tfrac{1}{2})^n$ (b) 10 folds (c) No — only approaches 0 in the limit

(a) $A_0 = 800$; halves each fold. $A_n = 800 \cdot (\frac{1}{2})^n = \frac{800}{2^n}$. (b) Need $\frac{800}{2^n} < 1$, i.e. $2^n > 800$. $2^9 = 512 < 800 < 1024 = 2^{10}$. So **10 folds**: $A_{10} = \frac{800}{1024} \approx 0.78$ cm². (c) No: $\frac{800}{2^n}$ is always **positive** for any finite $n$. The sequence converges to 0 but never reaches it — for any non-zero ε, there is an $n$ with $A_n < \varepsilon$, but no finite $n$ with $A_n = 0$.
6

**Mixed arithmetic/geometric.** A sequence has $u_1 = 6$. The differences between consecutive terms form a geometric sequence: $u_2 - u_1 = 4$, $u_3 - u_2 = 8$, $u_4 - u_3 = 16$, and so on. (a) Find $u_5$. (b) Find a closed form for $u_n$.

Answer

(a) $u_5 = 66$ (b) $u_n = 6 + 4(2^{n-1} - 1) = 2 + 2^{n+1}$

(a) Differences: $4, 8, 16, 32, \ldots$ (geometric, ratio 2). $u_2 = 6 + 4 = 10$; $u_3 = 10 + 8 = 18$; $u_4 = 18 + 16 = 34$; $u_5 = 34 + 32 = 66$. (b) $u_n = u_1 + \sum_{k=1}^{n-1}(\text{$k$th difference})$. The $k$th difference is $4 \cdot 2^{k-1}$. Sum: $\sum_{k=1}^{n-1} 4 \cdot 2^{k-1} = 4 \cdot \frac{2^{n-1} - 1}{2 - 1} = 4(2^{n-1} - 1)$. So $u_n = 6 + 4(2^{n-1} - 1) = 6 + 2^{n+1} - 4 = 2 + 2^{n+1}$. Check $u_5 = 2 + 2^6 = 2 + 64 = 66$ ✓.
7

**Card stacking.** Sienna builds a house of cards. Level 1 (top) uses 2 cards. Level 2 needs 5 cards. Level 3 needs 8 cards. Each level adds 3 more cards than the level above. (a) How many cards are needed for level $n$? (b) How many cards are needed to build all levels from 1 to 10? (c) Sienna has 200 cards. Including all levels, what is the largest house she can build?

Answer

(a) $u_n = 3n - 1$ (b) 155 cards (c) 11 levels (uses 187 cards)

(a) Arithmetic: $u_1 = 2$, $d = 3$. So $u_n = 2 + (n-1)\times 3 = 3n - 1$. (b) $S_{10} = \frac{10}{2}(u_1 + u_{10}) = 5(2 + 29) = 155$ cards. (c) Total for $n$ levels: $S_n = \frac{n}{2}(2 + (3n-1)) = \frac{n(3n+1)}{2}$. Solve $\frac{n(3n+1)}{2} \leq 200$, i.e. $n(3n+1) \leq 400$. - $n = 11$: $11 \times 34 = 374$ ≤ 400 ✓. - $n = 12$: $12 \times 37 = 444$ > 400 ✗. So 11 levels, using **$\frac{11 \times 34}{2} = 187$ cards** (13 left over).
8

**Identifying the type.** For each sequence, state whether it is arithmetic, geometric, or neither, and find a formula or rule for $u_n$. (a) $5, 9, 13, 17, 21, \ldots$ (b) $3, 6, 12, 24, 48, \ldots$ (c) $1, 4, 9, 16, 25, \ldots$ (d) $2, 5, 11, 23, 47, \ldots$

Answer

(a) Arithmetic, $u_n = 4n+1$ (b) Geometric, $u_n = 3 \cdot 2^{n-1}$ (c) Neither (square), $u_n = n^2$ (d) Neither, recursive $u_{n+1} = 2u_n + 1$

(a) Differences all 4 → arithmetic, $u_n = 4n + 1$. (b) Ratios all 2 → geometric, $u_n = 3 \cdot 2^{n-1}$. (c) Differences 3, 5, 7, 9 (not constant) and ratios not constant either, so neither arithmetic nor geometric. Pattern: $u_n = n^2$. (d) Differences 3, 6, 12, 24 — not arithmetic, but these differences themselves double. Ratios $\frac{5}{2}, \frac{11}{5}, \frac{23}{11}$ — not constant. Try recursive: $u_{n+1} = 2u_n + 1$: $2(2)+1=5$ ✓; $2(5)+1=11$ ✓; $2(11)+1=23$ ✓. So **recursive rule**: $u_{n+1} = 2u_n + 1$ with $u_1 = 2$. (Closed form: $u_n = 3 \cdot 2^{n-1} - 1$.)
9

**Two sequences meeting.** Sequence A: $u_n = 4n + 3$. Sequence B: starts at 47 and decreases by 2 each term. (a) Write a formula for sequence B. (b) Find the value of $n$ for which $A_n = B_n$. (c) What is the value of the common term?

Answer

(a) $B_n = 49 - 2n$ (b) $n = 7\tfrac{2}{3}$ — no integer solution; see working (c) No integer common term

(a) $B_1 = 47$, $d = -2$. So $B_n = 47 + (n-1)(-2) = 49 - 2n$. (b) Set $A_n = B_n$: $4n + 3 = 49 - 2n$. So $6n = 46$, $n = \frac{46}{6} = \frac{23}{3} \approx 7.67$. Since $n$ must be a positive integer for a sequence term, the sequences **do not share a term**. However the equation $4n + 3 = 49 - 2n$ has the (non-integer) solution $n = \frac{23}{3}$, at which both would equal $\frac{101}{3} \approx 33.67$. (c) The integer terms of A around this point: $A_7 = 31$, $A_8 = 35$. Terms of B: $B_7 = 35$, $B_8 = 33$. So $A_8 = B_7 = 35$ — the value 35 appears in **both** sequences, but at different positions. The common value is **35**.
10

**Recursive notation challenge.** A sequence is defined by $$u_1 = 1, \quad u_2 = 3, \quad u_{n+1} = u_n + 2 u_{n-1}.$$ (a) Find $u_3$, $u_4$, $u_5$, $u_6$. (b) Show that all terms are odd integers. (c) Calculate $\dfrac{u_{n+1}}{u_n}$ for $n = 1, 2, 3, 4, 5$. What do you notice?

Answer

(a) 5, 11, 21, 43 (b) See working (c) Ratios: 3, 1.67, 2.2, 1.91, 2.05 — approach 2

(a) $u_3 = u_2 + 2u_1 = 3 + 2 = 5$. $u_4 = u_3 + 2u_2 = 5 + 6 = 11$. $u_5 = u_4 + 2u_3 = 11 + 10 = 21$. $u_6 = u_5 + 2u_4 = 21 + 22 = 43$. (b) **Proof by induction.** Base cases: $u_1 = 1$ (odd), $u_2 = 3$ (odd). Inductive step: assume $u_n$ and $u_{n-1}$ are odd. Then $u_{n+1} = u_n + 2u_{n-1}$ = (odd) + (even) = odd. So by induction, all terms are odd. (c) Ratios: - $\frac{u_2}{u_1} = 3$ - $\frac{u_3}{u_2} = \frac{5}{3} \approx 1.67$ - $\frac{u_4}{u_3} = \frac{11}{5} = 2.2$ - $\frac{u_5}{u_4} = \frac{21}{11} \approx 1.91$ - $\frac{u_6}{u_5} = \frac{43}{21} \approx 2.05$ The ratios appear to **oscillate around 2 and converge to 2**. (In fact $u_n = \frac{1}{3}(2^{n+1} + (-1)^n)$, so ratio → 2.)
11

**Sum of squares puzzle.** Consider the sequence $1, 4, 9, 16, 25, \ldots$ of square numbers. (a) Sum the first five terms. (b) The formula $\displaystyle S_n = \frac{n(n+1)(2n+1)}{6}$ gives the sum of the first $n$ square numbers. Verify this for $n = 5$. (c) Find the sum of the first 20 square numbers. (d) Find the sum: $4 + 9 + 16 + 25 + \ldots + 400$.

Answer

(a) 55 (b) 55 ✓ (c) 2870 (d) 2869

(a) $1 + 4 + 9 + 16 + 25 = 55$. (b) Formula: $\frac{5 \times 6 \times 11}{6} = \frac{330}{6} = 55$ ✓. (c) $S_{20} = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = \mathbf{2870}$. (d) $4 + 9 + \ldots + 400 = 2^2 + 3^2 + \ldots + 20^2 = S_{20} - 1^2 = 2870 - 1 = \mathbf{2869}$.
12

**Aesthetics of growth.** A nautilus shell grows in a logarithmic spiral. Each chamber is $\phi$ times the previous, where $\phi = \frac{1+\sqrt{5}}{2} \approx 1.618$ (the golden ratio). The smallest chamber has area $1 \text{ mm}^2$. (a) Write a formula for the area $A_n$ of the $n$th chamber. (b) Estimate the area of the 8th chamber (3 s.f.). (c) The total area of the first $n$ chambers is given by a geometric sum. Find the total area of the first 8 chambers (3 s.f.).

Answer

(a) $A_n = \phi^{n-1}$ (b) $\approx 29.0$ mm² (c) $\approx 74.4$ mm²

(a) Geometric with $u_1 = 1$, $r = \phi$. So $A_n = \phi^{n-1}$. (b) $A_8 = \phi^7$. Compute: $\phi^2 \approx 2.618$; $\phi^4 \approx 6.854$; $\phi^7 \approx \phi^4 \cdot \phi^2 \cdot \phi \approx 6.854 \times 2.618 \times 1.618 \approx 29.0$. So $A_8 \approx \mathbf{29.0}$ mm². (c) Sum of geometric series: $S_8 = \dfrac{1(\phi^8 - 1)}{\phi - 1}$. With $\phi^8 \approx 46.98$ and $\phi - 1 \approx 0.618$: $S_8 \approx \dfrac{45.98}{0.618} \approx \mathbf{74.4}$ mm². **Connection to Fibonacci/golden ratio:** the ratios $\dfrac{F_{n+1}}{F_n}$ of consecutive Fibonacci numbers approach $\phi$, which is why nautilus shells, sunflower seed heads, and pine cones all exhibit Fibonacci-like spirals.