Applications: Sequences and Series

Each chair adds the same height, so the heights of stacks of 1, 2, 3, … chairs form a linear sequence h = dn + c. The two stacks we know give the common difference d first, and then c.

1. Let h cm be the height of a stack of n chairs. Going from 4 chairs to 10 chairs adds 6 chairs and 136 − 100 = 36 cm, so each chair adds d = 36 ÷ 6 = 6 cm.
2. So h = 6n + c. The stack of 4 is 100 cm tall, so 6 × 4 + c = 100, which gives c = 100 − 24 = 76. Check with the other stack: 6 × 10 + 76 = 136.
3. (a) h = 6n + 76. Each chair added raises the stack by 6 cm. A single chair is 6 + 76 = 82 cm tall, so 76 cm is the height of one chair less the 6 cm that every chair adds.
4. On the trolley, the top of a stack is 15 + 6n + 76 = 6n + 91 cm above the floor. It must be no more than 210 cm, so 6n + 91 ≤ 210, which gives 6n ≤ 119.
5. Divide both sides by 6: n ≤ 1196, which is 19.83 to 2 decimal places. The number of chairs is a whole number, so it is at most 19. (b) A stack on the trolley can have 19 chairs. Check: 19 chairs reach 6 × 19 + 91 = 205 cm, which passes under the doorway, and 20 chairs reach 211 cm, which does not.

answer(a) h = 6n + 76: each chair adds 6 cm, and 76 cm is one chair's height of 82 cm less that 6 cm; (b) 19 chairs

techniqueThe nth Term · Term-to-Term Rules

examsPSLE · O-Level · GCSE Higher · H2

When the second differences of a sequence are all the same, its nth term is quadratic, s = an² + bn + c, and the second difference is 2a. Taking an² away leaves a linear sequence, which gives b and c.

1. The first differences are 7 − 3 = 4, 12 − 7 = 5 and 18 − 12 = 6. The second differences are all 1, so the sequence is quadratic, s = an² + bn + c, with 2a = 1. So a = 12.
2. Subtract 12n² from each term. At n = 1, 2, 3, 4 this takes away 0.5, 2, 4.5 and 8, which leaves 2.5, 5, 7.5 and 10. These go up by 2.5 and start at 2.5, so they are 2.5n. So b = 2.5 and c = 0.
3. (a) s = 12n² + 52n. Check at n = 4: 12 × 16 + 52 × 4 = 8 + 10 = 18.
4. She reaches the bottom when s ≥ 120, that is 12n² + 52n ≥ 120. Multiply both sides by 2: n² + 5n − 240 ≥ 0. The roots of n² + 5n − 240 = 0 are n = −5 ± √25 + 9602 = −5 ± √9852, which are about 13.19 and −18.19. The negative root is rejected, because n counts seconds.
5. (b) At the end of the 13th second she has gone 84.5 + 32.5 = 117 m, which is short of 120 m. At the end of the 14th second she has gone 98 + 35 = 133 m. So she reaches the bottom during the 14th second.

answer(a) s = 12n² + 52n, that is s = 0.5n² + 2.5n; (b) during the 14th second

techniqueQuadratic Sequences

examsGCSE Higher

Each rebound is the same fraction of the one before, so the rebound heights form a geometric sequence. Its nth term is arⁿ⁻¹, where a is the first term and r is the common ratio.

1. The first rebound is 60% of 400 cm, which is 0.6 × 400 = 240 cm. Each rebound after that is 0.6 times the one before, so the heights form a geometric sequence with first term a = 240 and common ratio r = 0.6.
2. The height of the nth rebound is h_n = arⁿ⁻¹ = 240 × 0.6ⁿ⁻¹ cm.
3. (a) h₅ = 240 × 0.6⁴ = 240 × 0.1296 = 31.104. The 5th rebound is 31.1 cm high, correct to 3 significant figures.
4. For (b), solve 240 × 0.6ⁿ⁻¹ < 5. Divide both sides by 240: 0.6ⁿ⁻¹ < 5240. Take logarithms: (n − 1)log 0.6 < log 5240. Because log 0.6 is negative, dividing both sides by it reverses the inequality: n − 1 > log(5/240)log 0.6 ≈ 7.58.
5. So n > 8.58, and the first whole number above it is n = 9. (b) The 9th rebound is the first lower than 5 cm. Check: h₈ = 240 × 0.6⁷ ≈ 6.72 cm and h₉ = 240 × 0.6⁸ ≈ 4.03 cm.

answer(a) 31.1 cm; (b) the 9th rebound

techniqueGeometric Sequences

examsGCSE Higher · H2

The expression under the sigma gives the seats in row r, and the limits say which rows are added. A sum of (22 + 2r) splits into 22 added once for each row and 2 times the sum of r, where ∑_r=1ⁿ r = n(n+1)2.

1. Row r has (22 + 2r) seats, and r runs from 1 to 30. So the total is ∑_r=1³⁰ (22 + 2r).
2. Split the sum into ∑_r=1³⁰ 22 + 2∑_r=1³⁰ r. The first part adds 22 once for each of the 30 rows, which is 30 × 22 = 660. The second part is 2 × 30 × 312 = 930.
3. (a) The stand has 660 + 930 = 1590 seats. Check with the first and last rows: row 30 has 22 + 60 = 82 seats, and 302(24 + 82) = 15 × 106 = 1590.
4. The seats behind the reserved rows are ∑_r=11³⁰ (22 + 2r). This is the whole sum less rows 1 to 10, and ∑_r=1¹⁰ (22 + 2r) = 10 × 22 + 2 × 10 × 112 = 220 + 110 = 330.
5. (b) ∑_r=11³⁰ (22 + 2r) = 1590 − 330 = 1260 seats. Check: rows 11 to 30 are 20 rows, from 44 seats up to 82 seats, and 202(44 + 82) = 10 × 126 = 1260.

answer(a) ∑_r=1³⁰ (22 + 2r) = 1590 seats; (b) ∑_r=11³⁰ (22 + 2r) = 1260 seats

techniqueSigma Notation · The Sum of an Arithmetic Series

examsH2

The amounts saved each month rise by the same $5, so they form an arithmetic sequence, and the total is an arithmetic series: S_n = n2(2a + (n − 1)d), where a is the first term and d the common difference.

1. The monthly amounts 40, 45, 50, … form an arithmetic sequence with first term a = 40 and common difference d = 5. After n months her total, in dollars, is S_n = n2(2 × 40 + (n − 1) × 5).
2. (a) S₂₄ = 242(80 + 23 × 5) = 12 × (80 + 115) = 12 × 195 = 2340. After 24 months she has saved $2340.
3. For (b), simplify: S_n = n2(80 + 5n − 5) = n2(5n + 75). The total must be at least 3000, so n2(5n + 75) ≥ 3000. Multiply both sides by 2 and divide by 5: n² + 15n − 1200 ≥ 0.
4. The roots of n² + 15n − 1200 = 0 are n = −15 ± √225 + 48002 = −15 ± √50252, which are about 27.94 and −42.94. The negative root is rejected, because n counts months, so n ≥ 27.94.
5. (b) Her total first reaches $3000 at the end of month 28. Check: S₂₇ = 272(80 + 130) = 2835, which is short of $3000, and S₂₈ = 14 × (80 + 135) = 3010.

answer(a) $2340; (b) at the end of month 28, when her total is $3010

techniqueThe Sum of an Arithmetic Series · The nth Term

examsH2

Each round has 4 times as many shares as the round before, so the shares in each round form a geometric sequence. The total after n rounds is the sum of a geometric series, S_n = a(rⁿ − 1)r − 1.

1. Round n has 5 × 4ⁿ⁻¹ shares: 5, 20, 80, … This is a geometric sequence with a = 5 and r = 4, so the total after n rounds is S_n = 5(4ⁿ − 1)4 − 1 = 5(4ⁿ − 1)3.
2. (a) S₈ = 5(4⁸ − 1)3 = 5 × 655353 = 109225 shares.
3. For (b), solve 5(4ⁿ − 1)3 > 1000000. Multiply both sides by 3 and divide by 5: 4ⁿ − 1 > 600000, so 4ⁿ > 600001.
4. Take logarithms: n log 4 > log 600001, so n > log 600001log 4 ≈ 9.60. The first whole number above it is n = 10.
5. (b) The total first passes one million after 10 rounds. Check: S₉ = 5(262144 − 1)3 = 436905 and S₁₀ = 5(1048576 − 1)3 = 1747625. The model assumes every share keeps reaching 4 people who have not shared the video yet. As the video spreads, fewer such people are left, so real sharing slows down.

answer(a) 109225 shares; (b) after 10 rounds

techniqueThe Sum of a Geometric Series · Geometric Sequences

examsH2

Just after a dose, the body holds that dose and what is left of every earlier dose. Those amounts form a geometric series. When the common ratio r satisfies |r| < 1, the sums settle toward the sum to infinity, S_∞ = a1 − r.

1. Just after the nth dose, the body holds the new 200 mg, plus 20% of the dose before it, plus 20% of 20% of the one before that, and so on back to the first dose. In mg this is 200 + 200(0.2) + 200(0.2)² + … + 200(0.2)ⁿ⁻¹, a geometric series with a = 200 and r = 0.2.
2. Its sum is S_n = 200(1 − 0.2ⁿ)1 − 0.2 = 250(1 − 0.2ⁿ). For example, S₁ = 200, S₂ = 240 and S₃ = 248.
3. The common ratio is 0.2, and |0.2| < 1, so 0.2ⁿ gets closer and closer to 0 and the series has a sum to infinity: S_∞ = a1 − r = 2000.8 = 250.
4. (a) In the long run, the body holds 250 mg just after each dose.
5. The gap between S_n and 250 mg is 250 − 250(1 − 0.2ⁿ) = 250 × 0.2ⁿ. After the 3rd dose it is 250 × 0.008 = 2 mg, and after the 4th dose it is 250 × 0.0016 = 0.4 mg.
6. (b) The 4th dose is the first after which the amount, 249.6 mg, is within 1 mg of 250 mg.

answer(a) 250 mg; (b) after the 4th dose, when the amount is 249.6 mg

techniqueThe Sum to Infinity · The Sum of a Geometric Series

examsH2

A recurrence relation gives each term from the one before, so it needs a rule and a first term. A value that the rule leaves unchanged is found by putting the same letter on both sides.

1. Growing by 20% multiplies the number of trout by 1.2, and the catch then takes 300 away. So u_n+1 = 1.2u_n − 300, with u₀ = 2000.
2. Apply the rule one year at a time: u₁ = 1.2 × 2000 − 300 = 2100 and u₂ = 1.2 × 2100 − 300 = 2220.
3. (a) u₃ = 1.2 × 2220 − 300 = 2664 − 300 = 2364 trout.
4. A number L that stays the same satisfies L = 1.2L − 300. Subtract L from both sides: 0.2L = 300, so L = 1500. Check: 1.2 × 1500 − 300 = 1500.
5. Write the number of trout as 1500 + e_n, where e_n is the gap from 1500. Then 1500 + e_n+1 = 1.2(1500 + e_n) − 300 = 1500 + 1.2e_n, so e_n+1 = 1.2e_n: the gap grows by 20% each year, whether it is above 1500 or below.
6. (b) 1500 trout would stay the same. The second lake starts 100 below that: 1.2 × 1400 − 300 = 1380 after one year and 1.2 × 1380 − 300 = 1356 after two. Its gap below 1500 keeps growing, so it holds fewer trout every year and cannot keep up a catch of 300. Above 1500 the model has the trout growing without limit; a real lake runs short of food long before that, so the model is only used for the next few years.

answer(a) u_n+1 = 1.2u_n − 300 with u₀ = 2000, and u₃ = 2364 trout; (b) 1500 trout; the second lake holds 1356 trout after 2 years and keeps falling

techniqueRecurrence Relations · Term-to-Term Rules

examsGCSE Higher · H2

A proof by induction has two parts. The base case shows the statement is true for n = 1. The inductive step shows that if it is true for some n = k, it is also true for n = k + 1. Together they carry the statement to every positive whole number.

1. Let P(n) be the statement 1 + 3 + … + (2n − 1) = n². Base case, n = 1: the left side is 1 and the right side is 1² = 1, so P(1) is true.
2. Inductive step: assume that P(k) is true for some positive whole number k, that is, 1 + 3 + … + (2k − 1) = k².
3. Ring k + 1 uses 2(k + 1) − 1 = 2k + 1 tiles. So 1 + 3 + … + (2k − 1) + (2k + 1) = k² + 2k + 1 = (k + 1)², which is the statement P(k + 1).
4. (a) P(1) is true, and whenever P(k) is true, P(k + 1) is true. So by induction, 1 + 3 + … + (2n − 1) = n² for every positive whole number n: the first n rings use n² tiles.
5. (b) Twenty rings use 20² = 400 tiles and twelve rings use 12² = 144 tiles, so rings 13 to 20 need 400 − 144 = 256 more tiles. Check the first new ring: ring 13 uses 2 × 13 − 1 = 25 tiles, and 12² + 25 = 169 = 13².

answer(a) 1 + 3 + … + (2n − 1) = n² for every positive whole number n; (b) 400 − 144 = 256 more tiles

techniqueProof by Induction: Summing 1 to n · Sequences Worth Knowing

examsPSLE · GCSE Higher

To prove that an expression is divisible by 7 for every n, show it for n = 1, then assume it for n = k and write the expression for n = k + 1 in terms of the one for n = k. A number of days that is 1 more than a multiple of 7 is a whole number of weeks and 1 day.

1. Let P(n) be the statement that 8ⁿ − 1 is divisible by 7. Base case, n = 1: 8¹ − 1 = 7 = 7 × 1, so P(1) is true.
2. Inductive step: assume that P(k) is true for some positive whole number k, so 8^k − 1 = 7m for some whole number m.
3. Then 8^k+1 − 1 = 8 × 8^k − 1 = 8(8^k − 1) + 8 − 1 = 8(8^k − 1) + 7. Replace 8^k − 1 by 7m: this is 8 × 7m + 7 = 7(8m + 1), which is divisible by 7. So P(k + 1) is true.
4. (a) P(1) is true, and whenever P(k) is true, P(k + 1) is true. So by induction, 8ⁿ − 1 is divisible by 7 for every positive whole number n.
5. (b) With n = 4, 4096 − 1 = 4095 is divisible by 7, and 4095 = 7 × 585. So 4096 days is 585 whole weeks and 1 day more. After 585 whole weeks it is a Friday again, and 1 more day makes it a Saturday. The capsule is opened on a Saturday.

answer(a) 8ⁿ − 1 is divisible by 7 for every positive whole number n; (b) 4096 = 7 × 585 + 1, so it is opened on a Saturday

techniqueProof by Induction: Divisibility · Proof by Induction: Summing 1 to n