Mathematics

Answer Key

8.6 Patterns

Pack A — Answers

# Question Answer
1 Find the next two terms of the sequence 5, 8, 11, 14, … and describe the rule. 17, 20; rule: add 3 each time
2 The $n$-th term of a sequence is $T_n = 4n + 1$. Find $T_10$. 41
3 A pattern of dots has 4 dots in the 1st, 7 dots in the 2nd, 10 dots in the 3rd. (a) How many dots in the 4th? (b) Describe the rule. (a) 13 (b) Add 3 each time
4 Find the next term of: 1, 4, 9, … 16
5 Continue: $-3, -1, 1, 3, \ldots$ 5, 7
6 For the sequence with formula $T_n = 2n + 5$, find the first three terms. 7, 9, 11
7 A pattern of stars: 1st has 5, 2nd has 8, 3rd has 11. Find a rule. Add 3 each pattern. ($T_n = 3n + 2$.)
8 Find the common difference of the sequence 12, 8, 4, 0, −4, … $-4$
9 Verify whether 47 is a term of the sequence $T_n = 3n + 2$. Yes — $n = 15$.
10 Verify whether 40 is a term of the sequence $T_n = 3n + 2$. No — $n = 12.67$ (not a whole number).
11 Find a formula for the $n$-th term of the sequence 4, 7, 10, 13, … $T_n = 3n + 1$
12 Find the 20th term of $T_n = 5n - 2$. 98
13 A pattern of matchsticks: each new triangle adds 2 sticks. The first uses 3 sticks. Find a formula for $T_n$. $T_n = 2n + 1$
14 A sequence has first term 10 and common difference $-3$. Write a formula for $T_n$. $T_n = -3n + 13$ (or $13 - 3n$)
15 Find $n$ when $T_n = 71$ in the sequence $T_n = 4n - 5$. $n = 19$
16 A linear sequence has $T_3 = 11$ and $T_7 = 27$. (a) Find $d$. (b) Find $T_1$. (c) Write $T_n$. (a) $d = 4$ (b) $T_1 = 3$ (c) $T_n = 4n - 1$
17 A square pattern grows: 1st has 1 dot, 2nd has 4, 3rd has 9. Write a formula. $T_n = n^2$
18 A sequence: $1, 3, 6, 10, 15, \ldots$ (triangular numbers). Find $T_6$ and $T_7$. 21 and 28
19 A sequence of stair-step shapes: 1st uses 1 cube, 2nd uses 3, 3rd uses 6, 4th uses 10. Find the rule. $T_n = n(n+1)/2$
20 The Fibonacci sequence is 1, 1, 2, 3, 5, 8, … State the recursion and find $T_8$. $T_n = T_{n-1} + T_{n-2}$; $T_8 = 21$
21 A row of conference tables seats people. The pattern is $M = 6n + 2$. (a) How many people sit at 8 tables? (b) How many tables are needed to seat at least 100 people? (a) 50 (b) 17 tables
22 In Mr. Packer's arrangement, $a$ people sit on each side, $b$ at each end. Show $P = 2an + 2b$ and find $P$ when $a = 5$, $b = 3$, $n = 4$. $P = 46$
23 Patterns of matchsticks: 1st 3 sticks, 2nd 5, 3rd 7. (a) Formula for $T_n$. (b) Which pattern uses 51 sticks? (a) $T_n = 2n + 1$ (b) pattern 25
24 Find a formula and the 50th term of: 7, 11, 15, 19, … $T_n = 4n + 3$; $T_{50} = 203$
25 A growing pattern of squares uses 4 sticks for the first square, then adds 3 sticks per extra square in a row. (a) Formula for $T_n$. (b) How many squares can you make from 100 sticks? (c) How many sticks left over? (a) $T_n = 3n + 1$ (b) 33 squares; 0 left
26 Find a formula for $T_n$ in the sequence $-1, 2, 5, 8, 11, \ldots$ $T_n = 3n - 4$
27 The 4th term of a linear sequence is 17 and the 10th term is 41. Find a formula for $T_n$. $T_n = 4n + 1$
28 The sequence $5, 8, 13, 20, 29, \ldots$ has second differences 2. Find a formula for $T_n$. $T_n = n^2 + 4$
29 Find the sum of the first 10 terms of $T_n = 2n + 1$ (odd numbers starting from 3). 120
30 For the pattern $T_n = n^2 - n$, find $T_1, T_2, T_5, T_{10}$. 0, 2, 20, 90
31 The sequence of perfect squares is $1, 4, 9, 16, 25, \ldots$ (a) Formula for $T_n$. (b) Differences between consecutive terms. (c) Use this to find $T_{20}$ without squaring. (a) $T_n = n^2$. (b) Odd numbers 3, 5, 7, … (c) $T_{20} = 400$.
32 The 5th term of a linear sequence is 18 and the 12th term is 53. Find a formula, and find the smallest $n$ for which $T_n > 100$. $T_n = 5n - 7$; $n = 22$
33 For the sequence $1, 5, 12, 22, 35, \ldots$ (pentagonal numbers, formula $P_n = \tfrac{n(3n-1)}{2}$): verify the formula for $n = 1, 2, 3$, and find $P_{10}$. $P_1 = 1, P_2 = 5, P_3 = 12$; $P_{10} = 145$
34 Find the sum of all multiples of 3 between 1 and 100 (inclusive). 1683
35 A linear sequence has first term $a$ and common difference $d$. (a) Express $T_5$ in terms of $a$ and $d$. (b) If $T_5 = 23$ and $T_{12} = 65$, find $a$ and $d$. (a) $T_5 = a + 4d$ (b) $a = 1, d = 6$. Wait recompute → $a = 23 - 4 \times 6 = -1$. So $a = -1$, $d = 6$.
36 Two arithmetic sequences A: 5, 8, 11, … and B: 2, 7, 12, … . (a) Find formulas. (b) For which $n$ are the $n$-th terms equal? (a) $A_n = 3n + 2$, $B_n = 5n - 3$ (b) $n = 5/2$ — never (no whole-number solution)
37 A figurate-number pattern: the $n$-th hexagonal number is $H_n = n(2n-1)$. Find $H_6$ and $H_{10}$. $H_6 = 66, H_{10} = 190$
38 A geometric sequence has first term 3 and common ratio 2. (a) Find the first 5 terms. (b) Find a formula $T_n$. (c) Find $T_{10}$. (a) 3, 6, 12, 24, 48 (b) $T_n = 3 \times 2^{n-1}$ (c) 1536
39 A staircase pattern uses 1, 4, 9, 16, … cubes per layer. (a) State the formula for the $n$-th layer. (b) Find the total number of cubes in the first 5 layers. (c) Find a formula for the total $S_n = 1^2 + 2^2 + \ldots + n^2$ if you can. (a) $n^2$ (b) $1 + 4 + 9 + 16 + 25 = 55$ (c) $S_n = \tfrac{n(n+1)(2n+1)}{6}$
40 A sequence has $T_n = 2n^2 - n$. (a) Find $T_1, T_2, T_3$. (b) Show that the second differences are constant and find their value. (c) Is 119 a term of the sequence? (a) 1, 6, 15 (b) Second differences = 4 (c) No ($n \approx 8.1$)

Pack B — Answers

# Question Answer
1 Find the next two terms of the sequence 3, 9, 15, 21, … and describe the rule. 27, 33; rule: add 6 each time
2 The $n$-th term of a sequence is $T_n = 3n + 2$. Find $T_12$. 38
3 A pattern of dots has 4 dots in the 1st, 7 dots in the 2nd, 10 dots in the 3rd. (a) How many dots in the 4th? (b) Describe the rule. (a) 17 (b) Add 4 each time
4 Find the next term of: 2, 4, 8, … 16
5 Continue: $-3, -1, 1, 3, \ldots$ 5, 8
6 For the sequence with formula $T_n = 2n + 5$, find the first three terms. 2, 5, 8
7 A pattern of stars: 1st has 5, 2nd has 8, 3rd has 11. Find a rule. Add 4. ($T_n = 4n + 3$.)
8 Find the common difference of the sequence 12, 8, 4, 0, −4, … $-7$
9 Verify whether 47 is a term of the sequence $T_n = 3n + 2$. Yes — $n = 11$.
10 Verify whether 40 is a term of the sequence $T_n = 3n + 2$. Yes — $n = 7$.
11 Find a formula for the $n$-th term of the sequence 5, 9, 13, 17, … $T_n = 4n + 1$
12 Find the 20th term of $T_n = 5n - 2$. 5
13 A pattern of matchsticks: each new triangle adds 2 sticks. The first uses 3 sticks. Find a formula for $T_n$. $T_n = 3n + 1$
14 A sequence has first term 10 and common difference $-3$. Write a formula for $T_n$. $T_n = 5n + 3$
15 Find $n$ when $T_n = 71$ in the sequence $T_n = 4n - 5$. $n = 32$
16 A linear sequence has $T_3 = 11$ and $T_7 = 27$. (a) Find $d$. (b) Find $T_1$. (c) Write $T_n$. (a) $d = 4$ (b) $T_1 = 3$ (c) $T_n = 4n - 1$
17 A square pattern grows: 1st has 1 dot, 2nd has 4, 3rd has 9. Write a formula. Same.
18 A sequence: $1, 3, 6, 10, 15, \ldots$ (triangular numbers). Find $T_6$ and $T_7$. 36 and 45
19 A sequence of stair-step shapes: 1st uses 1 cube, 2nd uses 3, 3rd uses 6, 4th uses 10. Find the rule. Same.
20 The Fibonacci sequence is 1, 1, 2, 3, 5, 8, … State the recursion and find $T_8$. $T_{10} = 55$
21 A row of conference tables seats people. The pattern is $M = 6n + 2$. (a) How many people sit at 6 tables? (b) How many tables are needed to seat at least 80 people? (a) 38 (b) 13 tables
22 In Mr. Packer's arrangement, $a$ people sit on each side, $b$ at each end. Show $P = 2an + 2b$ and find $P$ when $a = 4$, $b = 2$, $n = 6$. $P = 52$
23 Patterns of matchsticks: 1st 3 sticks, 2nd 5, 3rd 7. (a) Formula for $T_n$. (b) Which pattern uses 99 sticks? (a) $T_n = 2n + 1$ (b) pattern 49
24 Find a formula and the 50th term of: 7, 11, 15, 19, … $T_n = 4n + 2$; $T_{50} = 202$
25 A growing pattern of squares uses 4 sticks for the first square, then adds 3 sticks per extra square in a row. (a) Formula for $T_n$. (b) How many squares can you make from 100 sticks? (c) How many sticks left over? (a) Same (b) 26 squares, 1 stick left
26 Find a formula for $T_n$ in the sequence $-1, 2, 5, 8, 11, \ldots$ $T_n = 3n - 8$
27 The 4th term of a linear sequence is 17 and the 10th term is 41. Find a formula for $T_n$. $T_n = 5n - 11$
28 The sequence $5, 8, 13, 20, 29, \ldots$ has second differences 2. Find a formula for $T_n$. $T_n = n^2 + 2$
29 Find the sum of the first 10 terms of $T_n = 2n + 1$ (odd numbers starting from 3). 168
30 For the pattern $T_n = n^2 - n$, find $T_1, T_2, T_5, T_{10}$. Same.
31 The sequence of perfect squares is $1, 4, 9, 16, 25, \ldots$ (a) Formula for $T_n$. (b) Differences between consecutive terms. (c) Use this to find $T_{20}$ without squaring. (a) $T_n = n^3$. (b) 7, 19, 37, … (c) $T_{10} = 1000$.
32 The 5th term of a linear sequence is 18 and the 12th term is 53. Find a formula, and find the smallest $n$ for which $T_n > 100$. $T_n = 5n - 8$; $n = 22$
33 For the sequence $1, 5, 12, 22, 35, \ldots$ (pentagonal numbers, formula $P_n = \tfrac{n(3n-1)}{2}$): verify the formula for $n = 1, 2, 3$, and find $P_{10}$. Same.
34 Find the sum of all multiples of 3 between 1 and 100 (inclusive). 1050
35 A linear sequence has first term $a$ and common difference $d$. (a) Express $T_5$ in terms of $a$ and $d$. (b) If $T_5 = 23$ and $T_{12} = 65$, find $a$ and $d$. (a) $T_4 = a + 3d$ (b) $a = 2, d = 3$
36 Two arithmetic sequences A: 5, 8, 11, … and B: 2, 7, 12, … . (a) Find formulas. (b) For which $n$ are the $n$-th terms equal? (a) $A_n = 4n - 1$, $B_n = 5n - 4$ (b) $n = 3$ — $A_3 = B_3 = 11$
37 A figurate-number pattern: the $n$-th hexagonal number is $H_n = n(2n-1)$. Find $H_6$ and $H_{10}$. Same.
38 A geometric sequence has first term 3 and common ratio 2. (a) Find the first 5 terms. (b) Find a formula $T_n$. (c) Find $T_{10}$. (a) 2, 6, 18, 54, 162 (b) $T_n = 2 \times 3^{n-1}$ (c) 39366
39 A staircase pattern uses 1, 4, 9, 16, … cubes per layer. (a) State the formula for the $n$-th layer. (b) Find the total number of cubes in the first 5 layers. (c) Find a formula for the total $S_n = 1^2 + 2^2 + \ldots + n^2$ if you can. Same.
40 A sequence has $T_n = 2n^2 - n$. (a) Find $T_1, T_2, T_3$. (b) Show that the second differences are constant and find their value. (c) Is 119 a term of the sequence? (a) 4, 14, 30 (b) Second diff = 6 (c) Test directly

Problems — Worked Solutions

1

**Conference tables (department Patterns and Formulae, January 2026).** Mr. Packer arranges square tables in a row. In Part One each table seats **3 people on each side and 1 person at each end**. (a) Draw or describe the arrangement for 1, 2, 3, 4 tables and complete a table of values. (b) Find an equation $M = \ldots$ for the number of people seated at $n$ tables. (c) Use it to find how many people can sit at 8 tables. (d) Verify by extending the table.

Answer

(a) 1→8, 2→14, 3→20, 4→26 (b) $M = 6n + 2$ (c) 50 people (d) Extending by +6 gives 8, 14, 20, 26, 32, 38, 44, 50 — matches.

(a) Each new table loses one end seat (becomes inner join) but gains 6 side seats — net +6. Values: 1 → 8, 2 → 14, 3 → 20, 4 → 26. (b) Common difference 6, first term 8: $M = 6n + 2$. Check $n=1$: $6+2 = 8$ ✓. (c) $M = 6(8) + 2 = 50$. (d) Adding 6 each time: 8, 14, 20, 26, 32, 38, 44, 50 ✓. Justification: 2 end seats + 6n side seats = $2 + 6n$.
2

**General Packer formula (Part 3).** In a more general arrangement, $a$ people sit on each side of a table and $b$ people at each end. (a) Build the formula for $P$, the total people seated, in terms of $a$, $b$, $n$ (number of tables). (b) Check your formula against Part One ($a = 3, b = 1$) and Part Two ($a = 4, b = 2$). (c) Mr. Packer wants to use 4 tables that seat 5 on each side and 3 at each end. He thinks he can seat 40 people. Use your formula to determine whether he is correct.

Answer

(a) $P = 2an + 2b$ (b) Part One: $P = 6n + 2$ ✓; Part Two: $P = 8n + 4$ (c) $P = 46$, not 40

(a) Each table has $a$ side seats on each of 2 sides → $2an$ total side seats. Only first & last tables have end seats → $2b$. So $P = 2an + 2b$. (b) Part One $(a=3, b=1)$: $P = 6n + 2$ ✓. Part Two $(a=4, b=2)$: $P = 8n + 4$. (c) $P = 2 \times 5 \times 4 + 2 \times 3 = 40 + 6 = 46$. Mr. Packer is **wrong** — 46 people, not 40.
3

**Linear sequence detective.** A sequence of pile-of-discs follows a linear pattern. The 4th term is 17 and the 10th term is 41. (a) Find the common difference. (b) Find the first term. (c) Write a formula for $T_n$. (d) Which term is equal to 101?

Answer

(a) 4 (b) 5 (c) $T_n = 4n + 1$ (d) $n = 25$

(a) From term 4 to term 10 is 6 steps with a change of 24. Common difference = $24/6 = 4$. (b) $T_1 + 3d = 17 \Rightarrow T_1 = 5$. (c) $T_n = 5 + 4(n-1) = 4n + 1$. (d) $4n + 1 = 101 \Rightarrow n = 25$.
4

**Square-grid patterns.** Diagram $n$ contains an $n \times n$ grid of small squares: - Diagram 1: 1 small square - Diagram 2: 4 small squares - Diagram 3: 9 small squares - Diagram 4: 16 small squares (a) Write a formula for $T_n$ — the number of small squares in diagram $n$. (b) Find the difference between consecutive diagrams. What do you notice? (c) Prove algebraically that $T_{n+1} - T_n = 2n + 1$. (d) Use this to find $T_{50}$ from $T_{49}$ without squaring 50.

Answer

(a) $T_n = n^2$ (b) Differences 3, 5, 7 — odd numbers (c) $(n+1)^2 - n^2 = 2n+1$ (d) $T_{50} = 2401 + 99 = 2500$

(a) $T_n = n^2$. (b) Differences 3, 5, 7 — consecutive odd numbers. (c) $(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$. (d) $T_{50} = T_{49} + (2 \times 49 + 1) = 2401 + 99 = 2500 = 50^2$ ✓.
5

**Consecutive integers.** Three consecutive integers have a sum of 96. (a) Let the smallest be $n$. Write expressions for the other two and an equation for the sum. (b) Solve to find the integers. (c) Show that the sum of any three consecutive integers is always a multiple of 3. (d) Is the sum of four consecutive integers always a multiple of 4? Justify algebraically.

Answer

(a) $n + (n+1) + (n+2) = 96$ (b) 31, 32, 33 (c) Sum $= 3n + 3 = 3(n+1)$ (d) No — sum $= 4n + 6 = 2(2n + 3)$, divisible by 2 not 4

(a) $n + (n+1) + (n+2) = 3n + 3 = 96$. (b) $n = 31$. Integers: 31, 32, 33. (c) Sum $= 3n + 3 = 3(n + 1)$ — always a multiple of 3, in fact $3 \times$ middle integer. (d) Sum of 4 consecutive integers $= 4n + 6 = 2(2n + 3)$. Since $2n + 3$ is odd, this is $2 \times \text{odd}$, divisible by 2 but **not by 4**.
6

**Matchstick triangle pattern.** A linear pattern of triangles built from matchsticks: - 1 triangle: 3 sticks - 2 triangles: 5 sticks - 3 triangles: 7 sticks - 4 triangles: 9 sticks (a) Find $T_n$. (b) How many triangles can be made from 81 sticks? (c) Comment on the sticks left over.

Answer

(a) $T_n = 2n + 1$ (b) 40 triangles (c) 0 sticks left over (uses all 81)

(a) Common difference 2, $T_1 = 3$ → $T_n = 2n + 1$. (b) Solve $2n + 1 \leq 81 \Rightarrow n \leq 40$. Make 40 triangles, using $2 \times 40 + 1 = 81$ sticks. (c) Zero sticks left over.
7

**Triangular numbers.** $T_n = 1 + 2 + 3 + \ldots + n$. (a) Find $T_1, T_2, T_3, T_4, T_5$. (b) Find a closed-form formula for $T_n$. (c) Find $T_{50}$. (d) Prove (e.g. by pairing the sum from both ends) that $T_n = n(n+1)/2$.

Answer

(a) 1, 3, 6, 10, 15 (b) $T_n = n(n+1)/2$ (c) $T_{50} = 1275$ (d) See working

(a) Cumulative sums: 1, 3, 6, 10, 15. (b) $T_n = n(n+1)/2$. (c) $T_{50} = 50 \times 51 / 2 = 1275$. (d) Pair: $1 + n$, $2 + (n-1)$, etc — each pair sums to $n + 1$. There are $n/2$ pairs (when $n$ even). Sum $= \tfrac{n(n+1)}{2}$. (For odd $n$, the same identity holds via the algebraic argument.)
8

**Fibonacci puzzle.** The Fibonacci sequence is defined by $F_1 = 1, F_2 = 1$, $F_{n+1} = F_n + F_{n-1}$. (a) Write the first 10 terms. (b) Find $F_{10}$. (c) Compute the ratio $F_{n+1}/F_n$ for $n = 5, 8, 10$. What do you notice as $n$ grows?

Answer

(a) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 (b) $F_{10} = 55$ (c) Ratios ≈ 1.6, 1.618, 1.618 — approaching the golden ratio $\varphi \approx 1.618$

(a) Each term is the sum of the two before: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. (b) $F_{10} = 55$. (c) $F_6/F_5 = 8/5 = 1.6$. $F_9/F_8 = 34/21 \approx 1.619$. $F_{11}/F_{10} = 89/55 \approx 1.618$. As $n \to \infty$, the ratio approaches the **golden ratio** $\varphi = (1 + \sqrt{5})/2 \approx 1.61803$.
9

**Sum-formula investigation.** The sum of an arithmetic sequence with first term $a$ and common difference $d$ over $n$ terms is $S_n = \tfrac{n(2a + (n-1)d)}{2}$ (also $S_n = \tfrac{n(\text{first} + \text{last})}{2}$). (a) Verify the formula for $1 + 2 + 3 + \ldots + 10$. (b) Use the formula to find the sum of $5 + 8 + 11 + \ldots + 32$. (c) The first $n$ odd numbers ($1 + 3 + 5 + \ldots$) have sum $n^2$. Verify for $n = 5$.

Answer

(a) 55 (b) 185 (c) $1+3+5+7+9 = 25 = 5^2$ ✓

(a) $a = 1, d = 1, n = 10$. $S_{10} = 10(2 + 9)/2 = 10 \times 11 / 2 = 55$. (b) Last term 32, $d = 3$. $n$: $5 + (n-1)3 = 32 \Rightarrow n = 10$. $S = 10(5 + 32)/2 = 185$. (c) $1 + 3 + 5 + 7 + 9 = 25 = 5^2$ ✓.
10

**Repeating-decimal investigation.** Recurring decimals can be converted to fractions using the "method of 9s": $0.\overline{a} = a/9$, $0.\overline{ab} = ab/99$, etc. (a) Convert $0.\overline{3}$, $0.\overline{35}$, $0.\overline{017}$ to fractions in simplest form. (b) Show algebraically that $0.\overline{9} = 1$. (c) Use the sequence of partial sums to argue that $0.999\ldots$ approaches 1.

Answer

(a) $\tfrac{1}{3}, \tfrac{35}{99}, \tfrac{17}{999}$ (b) Let $x = 0.\overline{9}$, then $10x - x = 9$, so $x = 1$ (c) Partial sums: 0.9, 0.99, 0.999, … → 1

(a) $\tfrac{3}{9} = \tfrac{1}{3}$; $\tfrac{35}{99}$; $\tfrac{17}{999}$. (b) $x = 0.\overline{9}$. $10x = 9.\overline{9}$. Subtract: $9x = 9$, $x = 1$. ∎ (c) Partial sums: $0.9, 0.9 + 0.09 = 0.99, 0.999, \ldots$ Each partial sum gets closer to 1 (difference: $0.1, 0.01, 0.001, \ldots$). The limit is exactly 1, formalising the algebra of part (b).
11

**Geometric pattern: paper folding.** A piece of paper is folded in half repeatedly. Each fold doubles the number of layers. (a) Find the number of layers after 1, 2, 3, 4, 5 folds. (b) Write a formula $L_n$ for the number of layers after $n$ folds. (c) Find $L_{10}$. (d) The Guinness world record for paper folds is 12. How many layers does that produce?

Answer

(a) 2, 4, 8, 16, 32 (b) $L_n = 2^n$ (c) 1024 (d) 4096

(a) $L = 2, 4, 8, 16, 32$. (b) Geometric, $L_n = 2^n$. (c) $L_{10} = 1024$. (d) $L_{12} = 4096$ layers.
12

**Mixed sequence puzzle.** Consider the sequence 2, 6, 12, 20, 30, 42, … (a) Find the next two terms. (b) Show the differences and second differences. What does that suggest about the formula? (c) Notice that $T_n = n(n+1)$. Verify for $n = 1, 2, 3, 4, 5$. (d) Find $T_{20}$.

Answer

(a) 56, 72 (b) First diffs 4, 6, 8, 10, 12; second diffs 2 (constant) → quadratic (c) $1 \times 2 = 2$, $2 \times 3 = 6$, etc. (d) 420

(a) Differences: 4, 6, 8, 10, 12, 14, 16. So next two: $42 + 14 = 56$; $56 + 16 = 72$. (b) First differences increase by 2 → second difference = 2 (constant). So the sequence is quadratic, $T_n = an^2 + bn + c$ with $2a = 2$, i.e. $a = 1$. (c) $T_n = n(n+1)$. $T_1 = 2, T_2 = 6, T_3 = 12, T_4 = 20, T_5 = 30$ ✓. (d) $T_{20} = 20 \times 21 = 420$.