Math Quiz 8.1-8.3 Review 2 Solution

HPC Quiz 8.1-8.3 Review #2 Name: Key Write the first five terms of the sequence. Assume it begins with 1. 1. aₙ = (1 + (-1)^n) / 2n a₁ = 0, a₂ = 1/2, a₃ = 0, a₄ = 1/4, a₅ = 0 2. aₙ = 1/√n a₁ = 1, a₂ = √2/2, a₃ = √3/3, a₄ = 1/2, a₅ = √5/5 Write an expression for the nth term of the sequence. Assume it begins with 1. 3. 1/2, 1/9, 1/28, 1/65, 1/126, ... numerator: always 1 denominator: n³ + 1 aₙ = 1 / (n³ + 1) 4. 2, 4, 2, 4, ... Alternates... typically means (-1)^n or (-1)^(n-1) will be involved. 4 = 3 + 1 5 = 3 + 2 so aₙ = 3 + (-1)^n 5. 28, 24, 20, 16, 12, ... arithmetic: d = -4, a₁ = 28 aₙ = a₁ + (n - 1)d aₙ = 28 + (n - 1)(-4) aₙ = -4n + 32 6. 1/3, -2/9, 4/27, -8/81, ... numerator: powers of 2 → (-2)^(n-1) denominator: powers of 3 → 3^n aₙ = (-2)^(n-1) / 3^n 7. 9, 6, 4, ... geometric: r = 2/3, a₁ = 9 aₙ = a₁ * r^(n-1) aₙ = 9 * (2/3)^(n-1) 8. 3/5, 4/9, 5/13, 6/17, ... numerator: arithmetic n + 2 denominator: different arithmetic 4n + 1 aₙ = (n + 2) / (4n + 1) Find the sum. 9. Σ(3i²) from i = 0 to 5 165 10. Σ(i² + 1) from i = 0 to 40 22,181 not arithmetic or geometric, so just add CALCULATOR PROGRAM! 11. Find the sum of the integers from 20 to 80 inclusive. 20 + 21 + 22 + ... + 79 + 80 arithmetic! n = 61 d = 1 a₁ = 20 Sₙ = (n/2)(a₁ + aₙ) = 3050 12. Determine the seating capacity of an auditorium with 30 rows of seats if there are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so   on. Seating capacity = Sum of all seats. S₃₀ = (30/2)(20 + 78) or (30/2)[2(20) + (30 - 1)(2)] = 1470 13. Which term of the arithmetic sequence -1, 4, 9, ... is 374? Finding the "n" term for an arithmetic sequence aₙ = 374 d = 5 a₁ = -1 aₙ = a₁ + (n - 1)d 374 = -1 + (n - 1)(5) 375 = (n - 1)(5) 75 = n - 1 n = 76 14. Find a₄₃ if a₃ = 13 and a₁₀ = 41. (arithmetic!) d = (41 - 13) / (10 - 3) = 4 13 = a₁ + (3 - 1)(4) 13 = a₁ + 8 a₁ = 5 a₄₃ = 5 + (43 - 1)(4) a₄₃ = 173 15. Give a recursive definition for the given sequence. a. 51, 43, 35, 27, ... arithmetic d = -8 a₁ = 51 aₖ₊₁ = aₖ - 8 b. 256, 64, 16, 4, ... geometric a₁ = 256 r = 1/4 aₖ₊₁ = (1/4)aₖ 16. Find the sum. If the series is arithmetic or geometric, use formulas. If the sequence is neither, you can use the sigma program on your calculator.   a. -6 - 2 + 2 + ... + 222 arithmetic a₁ = -6 d = 4 aₙ = a₁ + (n - 1)d 222 = -6 + (n - 1)(4) 228 = 4(n - 1) 57 = n - 1 n = 58 S₅₈ = (58/2)(-6 + 222) = 6264 b. 12 + 6 + 3 + 3/2 + ... + 3/256 geometric a₁ = 12 r = 1/2 n = 11 Sₙ = a₁(1 - rⁿ)/(1 - r) = 12(1 - (1/2)¹¹) / (1 - 1/2) = 6141/256 Write the sum using sigma notation. [Hint: are any of these arithmetic or geometric?] Then find the sum. 17. 5/(1+1) + 5/(1+2) + 5/(1+3) + ... + 5/(1+16) Σ(5/(1+n)) from n=1 to 16 = 64331/5274 18. 6 + 13 + 20 + 27 + ... + 97 Arithmetic! a₁ = 6, d = 7 97 = 6 + (n - 1)(7) 91 = 7(n - 1) 13 = n - 1 n = 14 S₁₄ = (14/2)(6 + 97) = 721 19. 3 - 9 + 27 - 81 + 243 - 729 Not arithmetic or geometric, so You can use your calculator! Σ(3ⁿ(-1)^(n-1)) from n=1 to 6 = -546 20. 7 - 11 + 15 - 19 + ... - 83 Not arithmetic or geometric... looks arithmetic but it alternates!! Ignore d and find n aₙ = a₁ + (n - 1)d 83 = 7 + (n - 1)(4) 76 = 4(n - 1) 19 = n - 1 n = 20 Σ((-1)^(n-1)(4n+3)) from n=1 to 20 = -40 Find the sum of the infinite geometric series, if possible. If it is not possible, explain why. 21. 4 + 6 + 9 + 13.5 + ... r = 6/4 = 9/6 = 13.5/9 = 1.5 Can't do, r > 1 22. 3 - 1 + 1/3 - 1/9 + ... r = -1/3 Since |r| < 1 a₁ = 3 6 / (1 - (-1/3)) = 9/2 23. Σ(6(2/3)ⁿ) from n=0 to ∞ r = 2/3 a₁ = 6(2/3)⁰ = 6 6 / (1 - 2/3) = 18 24. Σ(11(-4/3)^(n-1)) from n=1 to ∞ Can't do, r = -4/3, |r| > 1