Questions and Answers #1 Sequences

Questions and Answers Sheet Sequences Question #1 Give a recursive definition of the sequence {𝑎𝑛 }, 𝑛 = 1,2,3, if a) 𝑎𝑛 = 4𝑛 − 2 b) 𝑎𝑛 − 1 + (−1)𝑛 c) 𝑎𝑛 = 𝑛(𝑛 + 1) d) 𝑎𝑛 = 𝑛2 Answer: Step 1 a)Given a) 𝑎𝑛 = 4𝑛 − 2 Let us first determine the first value at 𝑛 = 1 𝑎1 = 4(1) − 2 = 4 − 2 = 2 Next we determine the recursive definition by writing 𝑎𝑛 in terms of 𝑎𝑛−1 𝑎𝑛 = 4𝑛 − 2 = 4𝑛 − 4 + 4 − 2 = [4(𝑛 − 1) − 2] + 4 = 𝑎𝑛−1 + 4 Thus the recursive definition is then: 𝑎1 = 2 𝑎𝑛 = 𝑎𝑛−1 + 4 when 𝑛 ≥ 2 Step 2 (b)Given: 𝑎𝑛 = 1 + (−1)𝑛 Let us first determine the first two values at 𝑛 = 1 and 𝑛 = 2 𝑎1 = 1 + (−1)1 = 1 − 1 = 0 𝑎2 = 1 + (−1)2 = 1 + 1 = 2 Next we determine the recursive definition by writing 𝑎𝑛 in terms of 𝑎𝑛−2 𝑎𝑛 = 1 + (−1)𝑛 = 1 + (−1)2 ⋅ (−1)𝑛−2 = 1 + 1(−1)𝑛−2 = 1 + (−1)𝑛−2 = 𝑎𝑛−2 Thus the recursive definition is then: 𝑎1 = 0 𝑎2 = 2 𝑎𝑛 = 𝑎𝑛−2 when 𝑛 ≥ 3 Step 3 (c) Given: 𝑎𝑛 = 𝑛(𝑛 + 1) Let us first determine the first value at 𝑛 = 1 𝑎1 = 1(1 + 1) = 1(2) = 2 Next we determine the recursive definition by writing 𝑎𝑛 in terms of 𝑎𝑛−1 𝑎𝑛 = 𝑛(𝑛 + 1) = 𝑛2 + 𝑛 = 𝑛2 − 𝑛 + 𝑛 + 𝑛 = 𝑛(𝑛 − 1) + 𝑛 + 𝑛 = (𝑛 − 1)𝑛 + 2𝑛 = 𝑎𝑛−1 + 2𝑛 Thus the recursive definition is then: 𝑎1 = 2 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 when 𝑛 ≥ 2 Step 4 (d) Given 𝑎𝑛 = 𝑛2 Let us first determine the first value at 𝑛 = 1 𝑎1 = 12 = 1 Next we determine the recursive definition by writing 𝑎𝑛 in terms of 𝑎𝑛−1 𝑎𝑛 = 𝑛2 = 𝑛2 − 2𝑛 + 2𝑛 − 1 + 1 = (𝑛2 − 2𝑛 + 1) + 2𝑛 − 1 = (𝑛 − 1)2 + 2𝑛 − 1 = 𝑎𝑛−1 + 2𝑛 − 1 Thus the recursive definition is then: 𝑎1 = 1 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 when 𝑛 ≥ 2 Result a) 𝑎1 = 2 and 𝑎𝑛 = 𝑎𝑛−1 + 4 when 𝑛 ≥ 2 b) 𝑎1 = 0 and 𝑎2 = 2 and 𝑎𝑛 = 𝑎𝑛−2 when 𝑛 ≥ 33 c) 𝑎1 = 2 and 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 when 𝑛 ≥ 2 d) 𝑎1 = 1 and 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 when 𝑛 ≥ 2 Question #2 Find a recurrence relation satisfied by this sequence. 𝑎𝑛 = 3 Answer: Given: 𝑎𝑛 = 3 Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 3 We note that each term is the same as the previous term (as all terms are equal to 3): 𝑎𝑛 = 𝑎𝑛−1 Thus a recurrence relation for 𝑎𝑛 is then: 𝑎0 = 3 𝑎𝑛 = 𝑎𝑛−1 Note: There are infinitely many different recurrence relations that satisfy any sequence. Question #3 Find a recurrence relation satisfied by this sequence. 𝑎𝑛 = 2𝑛 Answer: Given: 𝑎𝑛 = 2𝑛 Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 2(0) = 0 Let us similarly the next few terms as well: 𝑎1 = 2(1) = 2 = 𝑎0 + 2 𝑎2 = 2(2) = 4 = 𝑎1 + 2 𝑎3 = 2(3) = 6 = 𝑎2 + 2 We note that each term is the previous term increased by 2: 𝑎𝑛 = 𝑎𝑛−1 + 2 Thus a recurrence relation for a_{n} is then: 𝑎0 = 0 𝑎𝑛 = 𝑎𝑛−1 + 2 Note: There are infinitely many different recurence relations that satisfy any sequence. Question #4 Find a recurrence relation satisfied by this sequence. 𝑎𝑛 = 2𝑛 + 3 Answer: Given: 𝑎𝑛 = 2𝑛 + 3 Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 2(0) + 3 = 3 Let us similarly determine the next few terms as well: 𝑎1 = 2(1) + 3 = 5 = 𝑎0 + 2 𝑎2 = 2(2) + 3 = 7 = 𝑎1 + 2 𝑎3 = 2(3) + 3 = 9 = 𝑎2 + 2 We note that each term is the previous term increased by 2: 𝑎𝑛 = 𝑎𝑛−1 + 2 Thus a recurrence relation for 𝑎𝑛 is then 𝑎0 = 3 𝑎𝑛 = 𝑎𝑛−1 + 2 Note: There are infinitely many different recurence relations that satisfy any sequence. Question #5 Find a recurrence relation satisfied by this sequence. 𝑎𝑛 = 5𝑛 Answer: Given: 𝑎𝑛 = 5𝑛 Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 50 = 1 Let us similarly determine the next few terms as well: 𝑎1 = 51 = 5 = 5𝑎0 𝑎2 = 52 = 25 = 5𝑎1 𝑎3 = 53 = 125 = 5𝑎2 We note that each term is the previous term multiplied by 5: 𝑎𝑛 = 5𝑎𝑛−1 Thus a recurrence relation for 𝑎𝑛 is then: 𝑎0 = 1 𝑎𝑛 = 5𝑎𝑛−1 Note: There are infinitely many different recurence relations that satisfy any sequence. Question #6 Find a recurrence relation satisfied by this sequence. 𝑎𝑛 = 𝑛2 Answer: Given: 𝑎𝑛 = 𝑛2 Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 02 = 0 Let us similarly determine the next few terms as well: 𝑎1 = 12 = 1 = 𝑎0 + 2(1) − 1 𝑎2 = 22 = 4 = 𝑎1 + 2(2) − 1 𝑎3 = 32 = 9 = 𝑎2 + 2(3) − 1 𝑎4 = 42 = 16 = 𝑎3 + 2(4) − 1 𝑎5 = 52 = 25 = 𝑎4 + 2(5) − 1 𝑎6 = 62 = 36 = 𝑎5 + 2(6) − 1 We note that each term is the previous term increased by 2𝑛 − 1: 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 Thus a recurrence relation for 𝑎𝑛 is then: 𝑎0 = 0 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 Note: There are infinitely many different recurence relations that satisfy any sequence. Question #7 Find a recurrence relation satisfied by this sequence. 𝑎𝑛 = 𝑛2 + 𝑛 Answer: Given: 𝑎𝑛 = 𝑛2 + 𝑛 Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 02 + 0 = 0 Let us similarly determine the next few terms as well: 𝑎1 = 12 + 1 = 2 = 𝑎0 + 2(1) 𝑎2 = 22 + 2 = 6 = 𝑎1 + 2(2) 𝑎3 = 32 + 3 = 12 = 𝑎2 + 2(3) 𝑎4 = 42 + 4 = 20 = 𝑎3 + 2(4) 𝑎5 = 52 + 5 = 30 = 𝑎4 + 2(5) 𝑎6 = 62 + 6 = 42 = 𝑎5 + 2(6) We note that each term is the previous term increased by 2𝑛 − 1: 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 Thus a recurrence relation for 𝑎𝑛 is then: 𝑎0 = 0 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 1 Note: There are infinitely many different recurence relations that satisfy any sequence. Question #8 Find a recurrence relation satisfied by this sequence. 𝑎𝑛 = 𝑛 + (−1)𝑛 Answer: Given: 𝑎𝑛 = 𝑛 + (−1)𝑛 Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 0 + (−1)0 = 0 + 1 = 1 Let us similarly determine the next few terms as well: 𝑎1 = 1 + (−1)1 = 1 − 1 = 0 = 𝑎0 − 1 𝑎2 = 2 + (−1)2 = 2 + 1 = 3 = 𝑎1 + 3 𝑎3 = 3 + (−1)3 = 3 − 1 = 2 = 𝑎2 − 1 𝑎4 = 4 + (−1)4 = 4 + 1 = 5 = 𝑎3 + 3 𝑎5 = 5 + (−1)5 = 5 − 1 = 4 = 𝑎4 − 1 𝑎6 = 6 + (−1)6 = 6 + 1 = 7 = 𝑎5 + 3 We note that each term is the previous term increased by 3 if n is even: If n even: 𝑎𝑛 = 𝑎𝑛−1 + 3 We note that each term is the previous term decreased by 1 if n is odd: If n odd: 𝑎𝑛 = 𝑎𝑛−1 − 1 Thus a recurrence relation for 𝑎𝑛 is then: 𝑎0 = 1 a + 3 if n even an =  n −1 an −1 − 1 if n odd Note: There are infinitely many different recurence relations that satisfy any sequence. Question #9 Find a recurrence relation satisfied by this sequence. 𝑎𝑛 = 𝑛! Answer: Given: 𝑎𝑛 = 𝑛! Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 0! = 1 Let us first determine the first term by replacing n in the given expression for 𝑎𝑛 by 0: 𝑎0 = 0! = 1 Let us similarly determine the next few terms as well: 𝑎1 = 1! = 1 = 1𝑎0 𝑎2 = 2! = 2 = 2𝑎1 𝑎3 = 3! = 6 = 3𝑎2 𝑎4 = 4! = 24 = 4𝑎3 𝑎5 = 5! = 120 = 5𝑎4 𝑎6 = 6! = 720 = 6𝑎5 We note that each term is the previous term multiplied by n: 𝑎𝑛 = 𝑛𝑎𝑛−1 Thus a recurrence relation for 𝑎𝑛 is then: 𝑎0 = 1 𝑎𝑛 = 𝑛𝑎𝑛−1 Note: There are infinitely many different recurence relations that satisfy any sequence. Question #10 Show that the sequence 𝑎𝑛 is an solution of the recurrence relation 𝑎𝑛 = −3𝑎𝑛−1 + 4𝑎𝑛−2 if 𝑎𝑛 = 0 . Answer: Given: 𝑎𝑛 = 0 𝑛 = 0,1,2, … To prove: 𝑎𝑛 = −3𝑎𝑛−1 + 4𝑎𝑛−2 , 𝑛 ≥ 2 . Proof Replace n in 𝑎𝑛 = 0 by 𝑛 − 1: 𝑎𝑛−1 = 0 Replace n in 𝑎𝑛 = 0 by 𝑛 − 2: 𝑎𝑛−2 = 0 We will start from the expression −3𝑎𝑛−1 + 4𝑎𝑛−2 and prove that this term has to be equal to 𝑎𝑛 (when 𝑛 ≥ 2) . Let us use the two previous expressions derived for 𝑎𝑛−1 and 𝑎𝑛−2 −3𝑎𝑛−1 + 4𝑎𝑛−2 = −3(0) + 4(0) = 0 + 0 = 0 = 𝑎𝑛 Question #11 Show that the sequence 𝑎𝑛 is an solution of the recurrence relation 𝑎𝑛 = −3𝑎𝑛−1 + 4𝑎𝑛−2 if 𝑎𝑛 = 1 . Answer: Given: 𝑎𝑛 = 1 𝑛 = 0,1,2, … To prove: 𝑎𝑛 = −3𝑎𝑛−1 + 4𝑎𝑛−2 , 𝑛 ≥ 2. Proof Replace n in 𝑎𝑛 = 1 by 𝑛 − 1: 𝑎𝑛−1 = 1 Replace n in 𝑎𝑛 = 1 by 𝑛 − 2: 𝑎𝑛−2 = 1 We will start from the expression −3𝑎𝑛−1 + 4𝑎𝑛−2 and prove that this term has to be equal to 𝑎𝑛 (when 𝑛 ≥ 2) . Let us use the two previous expressions derived for 𝑎𝑛−1 and 𝑎𝑛−2 −3𝑎𝑛−1 + 4𝑎𝑛−2 = −3(1) + 4(1) = −3 + 4 = 1 = 𝑎𝑛 Question #12 Show that the sequence 𝑎𝑛 is an solution of the recurrence relation 𝑎𝑛 = −3𝑎𝑛−1 + 4𝑎𝑛−2 if 𝑎𝑛 = (−4)𝑛 . Answer: Given: 𝑎𝑛 = (−4)𝑛 𝑛 = 0,1,2, … To prove: 𝑎𝑛 = −3𝑎𝑛−1 + 4𝑎𝑛−2 , 𝑛 ≥ 2. Proof Replace n in 𝑎𝑛 = (−4)𝑛 by 𝑛 − 1: 𝑎𝑛−1 = (−4)𝑛−1 Replace n in 𝑎𝑛 = (−4)𝑛 by 𝑛 − 2: 𝑎𝑛−2 = (−4)𝑛−2 We will start from the expression −3𝑎𝑛−1 + 4𝑎𝑛−2 and prove that this term has to be equal to 𝑎𝑛 (when 𝑛 ≥ 2) . Let us use the two previous expressions derived for 𝑎𝑛−1 and 𝑎𝑛−2 −3𝑎𝑛−1 + 4𝑎𝑛−2 = −3(−4)𝑛−1 + 4(−4)𝑛−2 = −3 ∗ (−4) ∗ (−4)𝑛−2 + 4(−4)𝑛−2 Let us factor out (−4)𝑛−2 : = (−4)𝑛−2 (−3 ∗ (−4) + 4) = (−4)𝑛−2 (12 + 4) = (−4)𝑛−2 (16) = (−4)𝑛−2 (−4)2 = (−4)𝑛 = 𝑎𝑛 Question #13 Show that the sequence 𝑎𝑛 is an solution of the recurrence relation 𝑎𝑛 = −3𝑎𝑛−1 + 4𝑎𝑛−2 if 𝑎𝑛 = 2(−4)𝑛 + 3. Answer: Given: 𝑎𝑛 = 2(−4)𝑛 + 3 𝑛 = 0,1,2, … To prove: 𝑎𝑛 = −3𝑎𝑛−1 + 4𝑎𝑛−2 , 𝑛 ≥ 2 . Proof Replace n in 𝑎𝑛 = 2(−4)𝑛 + 3 by 𝑛 − 1: 𝑎𝑛−1 = 2(−4)𝑛−1 + 3 Replace n in 𝑎𝑛 = 2(−4)𝑛 + 3 by 𝑛 − 2: 𝑎𝑛−2 = 2(−4)𝑛−2 + 3 We will start from the expression −3𝑎𝑛−1 + 4𝑎𝑛−2 and prove that this term has to be equal to 𝑎𝑛 (when 𝑛 ≥ 2) . Let us use the two previous expressions derived for 𝑎𝑛−1 and 𝑎𝑛−2 −3𝑎𝑛−1 + 4𝑎𝑛−2 = −3(2(−4)𝑛−1 + 3) + 4(2(−4)𝑛−2 + 3) Use distributive property: = −6(−4)𝑛−1 − 9 + 8(−4)𝑛−2 + 12 Group the powers of -4: = (−6(−4)𝑛−1 + 8(−4)𝑛−2 ) − 9 + 12 = (−6(−4) ∗ (−4)𝑛−2 + 8(−4)𝑛−2 ) + 3 Let us factor out (−4)𝑛−2 : = (−4)𝑛−2 (−6 ∗ (−4) + 8) + 3 = (−4)𝑛−2 (24 + 8) + 3 = (−4)𝑛−2 ∗ 32 + 3 = (−4)𝑛−2 ∗ 16 ∗ 2 + 3 = (−4)𝑛−2 ∗ (−4)2 ∗ 2 + 3 = (−4)𝑛−2 ∗ 2 + 3 = 2(−4)𝑛 + 3 = 𝑎𝑛 Question #14 Given sequences 𝑆𝑛 = 3𝑛−4𝑛3 5𝑛2 +2 1 and 𝑡𝑛 = cos ( ). 𝑛 What conclusion we can make about each of the sequences? Answer: 𝑆𝑛 = 3 =𝑛 3𝑛 − 4𝑛3 5𝑛2 + 2 −4𝑛 2 5+ 2 𝑛 3𝑛 − 4𝑛 2 𝑛→∞ 𝑛→∞ 5+ 2 𝑛 0−∞ = = −∞ lim 𝑆𝑛 = lim 5+0 lim 𝑆𝑛 does not exists binitely 𝑛→∞ So that 𝑠𝑛 is a divergent sequence 1 𝑡𝑛 cos ( ) 𝑛 1 1 lim 𝑡𝑛 = lim cos ( ) = cos ( ) = cos(0) = 1 𝑛→∞ 𝑛→∞ 𝑛 ∞ So that t_n converges to 1 Question #15 Give the first six terms of the following sequences. You can assume that the sequences start with an index of 1. 1) An arithmetic sequence in which the first value is 2 and the common difference is 3. 2) A geometric sequence in which the first value is 27 and the common ratio is Answer: 1) The value 𝑢1 = 2 Common difference= 3 use the formula 𝑢𝑛 = 𝑢1 + (𝑛 − 1)𝑑 Replace 𝑢1 = 2, 𝑑 = 3 ⇒ 𝑢𝑛 = 2 + (𝑛 + 1)3 ⇒ 3𝑛 − 3 + 2 ⇒ 𝑢𝑛 = 3𝑛 − 1 𝑛 = 1 𝑢1 = 2 𝑛 = 2 𝑢2 = 5 𝑛 = 3 𝑢3 = 8 𝑛 = 4 𝑢4 = 11 𝑛 = 5 𝑢5 = 14 1 3 𝑛 = 6 𝑢6 = 17 2) a= first term= 27 1 𝛾 = 𝑐𝑜𝑚𝑚𝑜𝑛 𝑟𝑎𝑡𝑖𝑜𝑛 = 3 We know the formula 𝑎𝑛 = 𝑎𝛾 𝑛−1 Replace 𝑎 = 27, 𝛾 = 𝑎𝑛 𝑎𝑛 𝑎𝑛 𝑎𝑛 1 𝑛−1 = 27 ( ) 3 1 𝑛−1 = (3)3 ( ) 3 = 33 ⋅ 3−(𝑛−1) = 34−𝑛 1 3 1 1 3 9 𝑎1 = 27, 𝑎2 = 9, 𝑎3 = 3, 𝑎4 = 0, 𝑎5 = , 𝑎6 = Question #16 Write the first five terms of the sequence 2𝑛−1 { } 3𝑛+2 Answer: Substitute 𝑛 = 1, 2, 3, 4, 5 to find the first five sequences in the given sequence 2𝑛 − 1 { }= 3𝑛 + 2 2(1) − 1 1 { }= 3(1) + 2 5 2(2) − 1 3 { }= 3(2) + 2 8 2(3) − 1 5 { }= 3(3) + 2 11 2(4) − 1 7 { }= 3(4) + 2 14 2(5) − 1 9 { }= 3(5) + 2 17 Question #17 Do first five terms of sequence 1 − (−1)𝑛 { } 𝑛3 Answer: Given 1 − (−1)𝑛 { } 𝑛3 1 1 − (−1) 1 − (−1) { }= =2 13 1 2 1 − (−1) 1 − (1) { } =0 23 8 3 1 − (−1) 1 − (−1) 2 { } = 3 3 27 27 1 − (−1)4 1 − (−1) { } =0 43 64 1−(−1)5 1−(−1) { 53 } 125 = 2 125