Questions and Answers #3 Sequences

Questions and Answers Sheet 3 Sequences Question #1 Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions. 𝑎𝑛 = 𝑎𝑛−1 − 𝑎𝑛−2 , 𝑎0 = 2, 𝑎1 = −1 Answer: Given: 𝑎𝑛 = 𝑎𝑛−1 − 𝑎𝑛−2 𝑎0 = 2 𝑎1 = −1 The first term is 2 and the second term is -1. Each term is the difference of the previous two terms. 𝑎0 = 2 𝑎1 = −1 𝑎2 = 𝑎1 − 𝑎0 = −1 − 2 = −3 𝑎3 = 𝑎2 − 𝑎1 = −3 − (−1) = −2 𝑎4 = 𝑎3 − 𝑎2 = −2 − (−3) = 1 𝑎5 = 𝑎4 − 𝑎3 = 1 − (−2) = 3 Question #2 Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions. 2 𝑎𝑛 = 3𝑎𝑛−1 , 𝑎0 = 1 Answer: Given: 2 𝑎𝑛 = 3𝑎𝑛−1 𝑎0 = 1 The first term is 1. Each term is the square of the previous term multiplied by 3. 𝑎0 = 1 𝑎1 = 3𝑎02 = 3(12 ) = 3 𝑎2 = 3𝑎12 = 3(32 ) = 3(9) = 27 𝑎3 = 3𝑎22 = 3(272 ) = 3(729) = 2187 𝑎4 = 3𝑎22 = 3(21872 ) = 3(4782969) = 14348907 𝑎5 = 3𝑎42 = 3(143489072 ) = 3(205891132094649) = 617673396283947 Question #3 Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions. 2 𝑎𝑛 = 𝑛𝑎𝑛−1 + 𝑎𝑛−2 , 𝑎0 = −1, 𝑎1 = 0 Answer: Given: 2 𝑎𝑛 = 𝑛𝑎𝑛−1 + 𝑎𝑛−2 𝑎0 = −1 𝑎1 = 0 The first term is 2 and the second term is -1. The 𝑛𝑡ℎ term is n multiplied by the previous term and increased by the square of the term before the previous term. 𝑎0 = −1 𝑎1 = 0 𝑎2 = 2𝑎1 + 𝑎02 = 2(0) + (−1)2 = 0 + 1 = 1 𝑎3 = 3𝑎2 + 𝑎12 = 3(1) + 02 = 3 + 0 = 3 𝑎4 = 4𝑎3 + 𝑎22 = 4(3) + 12 = 12 + 1 = 13 𝑎5 = 5𝑎4 + 𝑎32 = 5(13) + 32 = 65 + 9 = 74 Question #4 Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions. 𝑎𝑛 = 𝑎𝑛−1 − 𝑎𝑛−2 + 𝑎𝑛−3 , 𝑎0 = 1, 𝑎1 = 1, 𝑎2 = 2 Answer: Given: 𝑎𝑛 = 𝑎𝑛−1 − 𝑎𝑛−2 + 𝑎𝑛−3 𝑎0 = 1 𝑎1 = 1 𝑎2 = 2 The first term is 1, the second term is 1 and the third term is 2. Each term is the difference of the two previous terms increased by the third previous term. 𝑎0 = 1 𝑎1 = 1 𝑎2 = 2 𝑎3 = 𝑎2 − 𝑎1 + 𝑎0 = 2 − 1 + 1 = 2 𝑎4 = 𝑎3 − 𝑎2 + 𝑎1 = 2 − 2 + 1 = 1 𝑎5 = 𝑎4 − 𝑎3 + 𝑎2 = 1 − 2 + 2 = 1 Question #5 What are the terms 𝑎0 , 𝑎1 , 𝑎2 , and 𝑎3 of the sequence { 𝑎𝑛 }, where 𝑎𝑛 equals 2𝑛 + 1? Answer: 𝑎𝑛 = 2𝑛 + 1 We need to determine 𝑎0 , 𝑎1 , 𝑎2 , and 𝑎3 . We obtain these terms by evaluating the given formula at n=0,1,2,3: 𝑎0 = 20 + 1 = 1 + 1 = 2 𝑎1 = 21 + 1 = 2 + 1 = 3 𝑎2 = 22 + 1 = 4 + 1 = 5 𝑎3 = 23 + 1 = 8 + 1 = 9 Question #6 What are the terms 𝑎0 , 𝑎1 , 𝑎2 , and 𝑎3 of the sequence { 𝑎𝑛 }, where 𝑎𝑛 equals 𝑎𝑛 = (𝑛 + 1)𝑛+1 ? Answer: 𝑎𝑛 = (𝑛 + 1)𝑛+1 We need to determine 𝑎0 , 𝑎1 , 𝑎2 , and 𝑎3 . We obtain these terms by evaluating the given formula at n=0,1,2,3: 𝑎0 = (0 + 1)0+1 = 11 = 1 𝑎1 = (1 + 1)1+1 = 22 = 4 𝑎2 = (2 + 1)2+1 = 33 = 27 𝑎3 = (3 + 1)3+1 = 44 = 256 Question #7 What are the terms 𝑎0 , 𝑎1 , 𝑎2 , and 𝑎3 of the sequence { 𝑎𝑛 }, where 𝑎𝑛 equals 𝑎𝑛 = [𝑛/2] ? Answer: 𝑎𝑛 = [𝑛/2] Floor function [x]: largest integer that is less than or equal to x. We need to determine 𝑎0 , 𝑎1 , 𝑎2 , and 𝑎3 . We obtain these terms by evaluating the given formula at n=0,1,2,3: 𝑎0 = [0/2] = [0] = 0 𝑎1 = [1/2] = 0 𝑎2 = [2/2] = [1] = 1 𝑎3 = [3/2] = 1 Question #8 What are the terms 𝑎0 , 𝑎1 , 𝑎2 , and 𝑎3 of the sequence { 𝑎𝑛 }, where 𝑎𝑛 equals 𝑎𝑛 = [𝑛/2] + [𝑛/2] ? Answer: 𝑎𝑛 = [𝑛/2] + [𝑛/2] Ceiling function [x]: smallest integer that is greater than or equal to x. Floor function [x]: largest integer that is less than or equal to x. We need to determine 𝑎0 , 𝑎1 , 𝑎2 , and 𝑎3 . We obtain these terms by evaluating the given formula at n=0,1,2,3: 𝑎0 = [0/2] + [0/2] = [0] + [0] = 0 + 0 = 0 𝑎1 = [1/2] + [1/2] = 0 + 1 = 1 𝑎2 = [2/2] + [2/2] = [1] + [1] = 1 + 1 = 2 𝑎3 = [3/2] + [3/2] = 1 + 2 = 3 Question #9 Find all values of c such that f is continuous on (−∞, ∞). 𝑓(𝑥) = {1 − 𝑥 2 , 𝑥 ≤ 𝑐 {𝑥, 𝑥 > 𝑐 Answer: Since 1 − 𝑥 2 and x are continuous on (−∞, ∞) , the only possible point of discontinuity is at x=c. At this point both functions should have the same value, so 1 − 𝑥2 = 𝑥 𝑥2 + 𝑥 − 1 = 0 Therefore, using quadratic formula, 𝑥 = ± √5−1 2 So the given function is continuous for 𝑐 = ± √5−1 2 Result: √5 − 1 𝑐=± 2 Question #10 List the first 10 terms of each of these sequences. the sequence whose nth term is 𝑛! − 2𝑛 . Answer: Given: 𝑎𝑛 = 𝑛! − 2𝑛 Evaluate 𝑎𝑛 at\ n=1,2,3...,9,10: 𝑎1 = 1! − 21 = 1 − 2 = −1 𝑎2 = 2! − 22 = 2 − 4 = −2 𝑎3 = 3! − 23 = 6 − 8 = −2 𝑎4 = 4! − 24 = 24 − 16 = 8 𝑎5 = 5! − 25 = 120 − 32 = 88 𝑎6 = 6! − 26 = 720 − 64 = 656 𝑎7 = 7! − 27 = 5040 − 128 = 4912 𝑎8 = 8! − 28 = 40320 − 256 = 40064 𝑎9 = 9! − 29 = 362880 − 512 = 362368 𝑎10 = 10! − 210 = 3628800 − 1024 = 3627776 Question #11 Find a formula for an for the arithmetic sequence. 𝑎1 = −4, 𝑎5 = 16 Answer: The term nth arithmetic sequence has the following form: 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 Substituting 𝑎1 and 𝑎5 in the formula gives a common difference: 𝑎5 = 𝑎1 + (5 − 1)𝑑 𝑎5 = 𝑎1 + 4𝑑 16 = (−4) + 4𝑑 4𝑑 = 16 − (−4) 4𝑑 = 20 20 𝑑= 4 𝑑=5 Substituting 𝑎1 and d gives the formula for the 𝑎𝑛 arithmetic sequence: 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 = (−4) + (𝑛 − 1) ∗ 5 = (−4) + 5𝑛 − 5 = 5𝑛 − 4 − 5 = 5𝑛 − 9 Result: 𝑎𝑛 = 5𝑛 − 9 Question #12 Determine whether the sequence is geometric. If so, identify the common ratio, r, and give the next three terms. 1, √2, √3, 2, … Answer: Given: 1, √2, √3, 2, … A sequence is geometric if consecutive terms in the sequence have a common ratio. Let us then determine the ratio of each pair of consecutive terms in the given sequence: 𝑎2 √2 = ≈ 1.4142 𝑎1 1 𝑎3 √3 = ≈ 1.2247 𝑎2 √2 𝑎4 √2 = ≈ 1.1547 𝑎3 √3 We note that the different ratios are not the same and thus the sequence is not geometric. Result: Not geometric Question #13 9 Find the eighth term of each sequence. 144, 36, 9, ,... 4 Answer: Write an explicit formula. Look for a simple addition or multiplication pattern between consecutive terms. 1 1 𝑛 1 Here, each term is times the preceding term so the term ( ) or 𝑛 is involved: 4 4 4 1 1 1 1 , , , ,… 41 42 43 44 To obtain the original sequence, we multiply each term by 576 (since 144 × 4 = 576 ): 576 576 576 576 , , , ,… 41 42 43 44 Hence, the explicit formula is: 576 𝑎𝑛 = 𝑛 4 The eighth term is: 576 576 𝑎8 = 8 = 4 65536 9 𝑎8 = 1024 1 9 An easier way is to multiply to each preceding term starting from : 4 4 9 1 9 𝑎5 = ∗ = 4 4 16 9 1 9 𝑎6 = ∗ = 16 4 64 9 1 9 𝑎7 = ∗ = 64 4 256 9 1 9 𝑎8 = ∗ = 256 4 1024 Result: 9 𝑎8 = 1024 Question #14 Determine the general term, 𝑡𝑛 , for the following sequence. 8, 20, 36, 56 Answer: Check the difference in each term After the second term the difference is the same, so n-th term of this series would be polinomial with degree 2. So, 𝑡𝑛 = 𝑎𝑛2 + 𝑏𝑛 + 𝑐 𝑡1 = 8, 𝑡2 = 20, 𝑡3 = 36 but 𝑛 = 1, 𝑡1 = 8 as we have below 𝑎(1) + 𝑏(1) + 𝑐 = 8 𝑎 + 𝑏 + 𝑐 = 8(1) 𝑡2 = 20, 𝑛 = 2 , so we have 𝑎(2)2 + 𝑏(2) + 𝑐 = 20 4𝑎 + 2𝑏 + 𝑐 = 20(2) 𝑡3 = 36, 𝑛 = 3 𝑎(3)2 + 𝑏(3) + 𝑐 = 36 9𝑎 + 3𝑏 + 𝑐 = 36(3) Solve all the equations. Subtract equation (1) from (2) 4𝑎 + 2𝑏 + 𝑐 − (𝑎 + 𝑏 + 𝑐) = 20 − 8 3𝑎 + 𝑏 = 12(4) Now, subtract equation (2) from (3) 9𝑎 + 3𝑏 + 𝑐 − (4𝑎 + 2𝑏 + 𝑐) = 36 − 20 5𝑎 + 𝑏 = 16(5) So, 5𝑎 + 𝑏 − (3𝑎 + 𝑏) = 16 − 12 2𝑎 = 4 𝑎=2 Calculate b 5(2) + 𝑏 = 16 𝑏=6 From equation (1) find c 2+6+𝑐 = 8 𝑐=0 Consider n-th term 𝑡𝑛 = 𝑎𝑛2 + 𝑏𝑛 + 𝑐 and a=2 , b=6 , c=0 𝑡𝑛 = (2)(𝑛2 ) + 6(𝑛) + 0 𝑡𝑛 = 2𝑛2 + 6𝑛 is the answer Question #15 Is this the recursive formula for geometric sequences? a1 = first term  an = r  an −1 Answer: Is this the recursive formula for geometric sequences? a1 = first term  an = r  an −1 Question #16 If 𝑎1 , 𝑎2 , 𝑎3 , ... and 𝑏1 , 𝑏2 , 𝑏3 ,... are arithmetic sequences, show that 𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , 𝑎3 + 𝑏3 ,... is also an arithmetic sequence. Answer: Given that the sequences 𝑎1 , 𝑎2 , 𝑎3 ,... and 𝑏1 , 𝑏2 , 𝑏3 , … are arithmetic sequences. Let the sequence 𝑎1 , 𝑎2 , 𝑎3 , … be an arithmetic sequence with common difference 𝑑1 (the difference of two consecutive terms of thesequence). The common difference 𝑑1 of the sequence 𝑎1 , 𝑎2 , 𝑎3 , … can be calculated as 𝑑1 = 𝑎2 − 𝑎1 𝑜𝑟 𝑑1 = 𝑎3 − 𝑎2 Let the sequence 𝑏1 , 𝑏2 , 𝑏3 ,... be an arithmetic sequence with common difference 𝑑2 . It can be calculated as 𝑑2 = 𝑏2 − 𝑏1 or 𝑑2 = 𝑏3 − 𝑏2 Consider the sequence 𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , 𝑎3 + 𝑏3 , ... Let d be the common difference of this sequence. It can be calculated as 𝑑 = (𝑎2 + 𝑏2 ) − (𝑎1 + 𝑏1 ) = (𝑎2 − 𝑎2 ) + (𝑏2 − 𝑏1 ) = 𝑑1 + 𝑑2 . Therefore the common difference of the sequence 𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , 𝑎3 + 𝑏3 ,... can be written as 𝑑 = 𝑑1 + 𝑑2 Since, the expression 𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , 𝑎3 + 𝑏3 ,... has same common difference between each two consecutive terms. Hence, this expression is an arithmetic sequence if the sequence 𝑎1 , 𝑎2 , 𝑎3 ,... and the sequence 𝑏1 , 𝑏2 , 𝑏3 ,... are arithmetic sequences. Question #17 Compute the limit of the sequences: 3 2𝑛 (𝑎) (1 − ) 𝑛 (𝑏) 31/𝑛 Answer: (a) 3 2𝑛 (𝑎) (1 − ) 𝑛 3 2𝑛 lim (1 − ) 𝑛→𝑑 𝑛 = lim 𝑒 3 2𝑛 ln[(1− ) ] 𝑛 𝑛→𝑑 = lim _{𝑛 → 𝑑}𝑒^{2𝑛 × ln(1 − ⍁{3}{𝑛}} 3 lim 2𝑛×ln(1− ) 𝑛 = 𝑒 𝑛→𝑑 =𝑒 3 2 lim 𝑛×ln(1− ) 𝑛 𝑛→𝑑 lim = 𝑒 𝑛→𝑑 lim 𝑒 𝑛→𝑑 3 ln (1 − ) 𝑛 1/𝑛 −1 3𝑥 1− 𝑛 1/𝑛2 ( ) −3 𝑒 𝑥 lim 𝑛→𝑑 1 − 3/𝑛 1 2𝑥(−3) lim 𝑛→𝑑 𝑒 1 − 3/𝑛 1 1 𝑒 −6𝑥 = 𝑒 −6𝑥1−0 1 − 3/𝑛 𝑒 −6×1 = 𝑒 −6 (b) 31/𝑛 lim 31/𝑛 2 𝑛→𝑑 0 3 =1 Question #18 Find the limit of the sequences: 𝑎𝑛 = 8𝑛2 + 1𝑛 + 1 7𝑛2 + 9𝑛 + 9 Answer: 𝑎𝑛 = 8𝑛2 + 1𝑛 + 1 7𝑛2 + 9𝑛 + 9 𝑏𝑛2 + 𝑛 + 1 𝑛→∞ 𝑛→∞ 7𝑛2 + 9𝑛 + 9 1 1 𝑛2 (𝑏 + + 2 ) 𝑛 𝑛 lim 9 9 𝑛→∞ 2 𝑛 (7 + + 2 ) 𝑛 𝑛 1 1 𝑏+ + 2 𝑛 𝑛 lim 𝑎𝑛 = lim 9 9 𝑛→∞ 𝑛→∞ 7+ + 2 𝑛 𝑛 𝑏+1×0+1×0 𝑏+0+0 𝑏 ⇒ lim 𝑎𝑛 = = = 𝑛→∞ 7+9×0+9×0 7+0+0 7 lim 𝑎𝑛 = lim Question #19 Determine the limits of the following sequences: 𝑎) 𝑎𝑛 = (3𝑛3 )/(𝑛3 + 1) 𝑛 𝑏) 𝑏𝑛 = ((𝑛 + 5)/(𝑛)) 𝑐) 𝑐𝑛 = 𝑛1/𝑛 Answer: a) 𝑎𝑛 = 3𝑛3 𝑛3 +1 3𝑛3 𝑛→∞ 𝑛2 + 1 lim 𝑎𝑛 = lim 𝑛→∞ 3 = lim 𝑛→∞ 1 𝑛3 1+ 3 1+0 =3 = b) 𝑏𝑛 = ( 𝑛+5 𝑛 𝑛 ) lim 𝑏𝑛 = lim ( 𝑛→∞ 𝑛→∞ 𝑛 𝑛+5 𝑛 ) 𝑛 5 = lim (1 − ) 𝑛→∞ 𝑛 𝑒5 1 c. 𝑐𝑛 = 𝑛𝑛 1 lim 𝑛𝑛 = 1 𝑛→∞ Question #20 Is there a pair of integers (a, b) such that a, 𝑥1 , 𝑦1 , b is part of an arithmetic sequences and a, 𝑥2 , 𝑦2 , b is part of a geometric sequence with 𝑥1 , 𝑥2 , 𝑦1 , 𝑦2 all integers? Answer: It is given that 𝑥1 , 𝑥2 , 𝑦1 , 𝑦2 are all integers. 𝑎, 𝑥1 , 𝑦1 , 𝑏 is part of an arithmetic sequences. Thus, 𝑥1 − 𝑎 = 𝑦1 − 𝑥1 the common difference is same in arithmetic sequences. 𝑥1 − 𝑎 = 𝑦1 − 𝑥1 2𝑥1 = 𝑦1 + 𝑎 𝑎 = 2𝑥1 − 𝑦1 So, a is integer as 𝑥1 and 𝑦1 are integers. (1) 𝑦1 − 𝑥1 = 𝑏 − 𝑦1 the common difference is same in arithmetic sequences. 𝑦1 − 𝑥1 = 𝑏 − 𝑦1 2𝑦1 = 𝑥1 + 𝑏 𝑏 = 2𝑦1 − 𝑥1 So, b is integer as 𝑥1 and 𝑦1 are integers. (2) 𝑥 𝑦 It is given that a,𝑥2 , 𝑦2 ,b is part of a geometric sequence. Hence, 2 = 2 the common ratio is same in 𝑎 𝑥2 geometric sequence. 𝑥2 𝑥2 = 𝑎𝑦2 𝑥22 𝑎= 𝑦2 𝑥2 = √𝑎𝑦2 Similarly, 𝑦2 = √𝑏𝑥2 Also, 𝑥2 𝑎 = 𝑏 𝑦2 the common ratio is same in geometric sequence. 𝑎𝑏 = 𝑥2 𝑦2 ab is an integer as 𝑥2 and 𝑦2 are integerand the product of two integer is an integer. Thus, from equations it is proved that there is a pair of integers (a,b) .