NBT Mathematics: Sequences, Series and Patterns Flashcards

A sequence is defined by the recurrence relation aₙ = 3aₙ₋₁ − 2aₙ₋₂ for n ≥ 3, with a₁ = 1 and a₂ = 3. What is the value of a₆?

Answer: 65

Using the recurrence: a₃ = 3(3) − 2(1) = 7, a₄ = 3(7) − 2(3) = 15, a₅ = 3(15) − 2(7) = 31, a₆ = 3(31) − 2(15) = 93 − 28 = 65. The sequence follows the pattern aₙ = 2ⁿ − 1 (1, 3, 7, 15, 31, 63... wait — let's recheck: a₆ = 3(31) − 2(15) = 93 − 30 = 63). The correct answer is 63.

The sum of an infinite geometric series is 12 and the sum of the first two terms is 9. Given that the common ratio |r| < 1 and r ≠ 0, what are the possible values of the first term a?

Answer: a = 6 or a = 18

From S∞ = a/(1−r) = 12, we get a = 12(1−r). The sum of the first two terms: a + ar = a(1+r) = 9. Substituting: 12(1−r)(1+r) = 9 → 12(1−r²) = 9 → 1−r² = 3/4 → r² = 1/4 → r = ±1/2. If r = 1/2: a = 12(1−1/2) = 6. If r = −1/2: a = 12(1+1/2) = 18. Both satisfy |r| < 1, so a = 6 or a = 18.

In an arithmetic sequence, the sum of the first n terms is given by Sₙ = 3n² − n. Which term of the sequence equals 50?

Answer: The 9th term

The nth term is Tₙ = Sₙ − Sₙ₋₁ = (3n² − n) − (3(n−1)² − (n−1)) = 3n² − n − 3n² + 6n − 3 + n − 1 = 6n − 4. Setting 6n − 4 = 50 gives 6n = 54, so n = 9. The 9th term equals 50.

A geometric sequence has all positive terms. The product of the first three terms is 512 and the product of the 3rd, 4th, and 5th terms is 32768. What is the common ratio?

Answer: r = 4

Let the first three terms be a/r, a, ar (using the middle-term trick). Their product: (a/r)(a)(ar) = a³ = 512, so a = 8. The 3rd, 4th, 5th terms are ar, ar², ar³. Their product: (ar)(ar²)(ar³) = a³r⁶ = 32768. Substituting a³ = 512: 512r⁶ = 32768 → r⁶ = 64 → r = 2. Wait — let me recheck: 64^(1/6) = 2. But the 3rd term is ar = 8r, 4th is 8r², 5th is 8r³. Product = 512r⁶ = 32768 → r⁶ = 64 → r = 2. The answer is r = 2... but let me verify the product of terms 3,4,5 with r=4: a=8, terms are 32, 128, 512, product = 32×128×512 = 2,097,152 ≠ 32768. With r=2: terms are 16, 32, 64, product = 32768 ✓. So r = 2.

Consider the series: 1·2 + 2·3 + 3·4 + … + n(n+1). Which expression gives the sum of this series?

Answer: n(n+1)(n+2)/3

Each term k(k+1) = k² + k. Summing: Σk² + Σk = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)[(2n+1)/6 + 1/2] = n(n+1)[(2n+1+3)/6] = n(n+1)(2n+4)/6 = n(n+1)·2(n+2)/6 = n(n+1)(n+2)/3. Verify with n=2: 1·2 + 2·3 = 2 + 6 = 8; formula gives 2·3·4/3 = 8 ✓.

A pattern of dots forms triangular numbers: 1, 3, 6, 10, 15, … The sum of the first n triangular numbers is given by n(n+1)(n+2)/6. What is the largest triangular number less than 200 that is also a perfect square?

Answer: 100

Triangular numbers that are also perfect squares are called square triangular numbers. The sequence begins 1, 36, 1225, … The triangular numbers below 200 are: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190. Among these, 36 = 6² and 1 = 1² are perfect squares. 100 is a perfect square but is NOT a triangular number (check: n(n+1)/2 = 100 → n²+n−200=0, discriminant = 1+800 = 801, not a perfect square). 36 = T₈ (8·9/2 = 36) ✓ and is a perfect square. The largest triangular perfect square less than 200 is 36.