Math 115 Spring 2014 Written Homework 3 Due Wednesday, February 19

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1 Math 11 Spring 01 Written Homework 3 Due Wednesday, February 19 Instructions: Write complete solutions on separate paper (not spiral bound). If multiple pieces of paper are used, they must be stapled with your name and lecture written on each page. In general, answers with no supporting work will not receive full credit. Please review the Course Information document for more complete instructions. 1. For each sequence, classify it as arithmetic or geometric, find a function a(n) that describes the sequence, and determine the limit of the sequence. (a) k 3, k 7, k 11, k 1, k 19,... where k is any real number. Solution: The sequence is arithmetic with a common difference of as = k 7 (k 3) = (k 11) (k 7) =.... The generating function is a n := k 3 + (n 1)( ) and the sequence decreases without bound as d < 0, so lim a n =. (b),,,,... Solution: The sequence is geometric with common ratio as = = = =.... The generating function is a n := ( ) n 1 = n. Since > 1 and a 1 > 0, lim a n =. (c) 7, 9, 3, 1, 1,... 3 Solution: The sequence is geometric with common ratio 1 3 Since 1 < 1 < 1, lim a 3 n = 0. as 1 3 = 7 9 = 9 3 = 3 1 =.... (d) 3, 7,, 9,,... Solution: The sequence is arithmetic with a common difference of 1 as 1 = 7 3 = 7 =.... The generating function is a n := 3 + (n 1)( 1 ) and the sequence increases without bound as d > 0, so lim a n =. (e) 7,, 3, 8, 13,... Solution: The sequence is arithmetic with a common difference of as = ( 7) = 3 ( ) = 8 3 =.... The generating function is a n := 7 + (n 1)() and the sequence increases without bound as d > 0, so lim a n =. (f),,, 10,... Solution: The sequence is geometric with common ratio as = Since r < 1, lim a n D.N.E. = = 10 =....

2 . Determine the function that generates the finite sequence 7, 1, 17,,..., 117 and determine how many terms are in the sequence. Solution: First, we should find the simplest pattern. This means checking if the sequence is arithmetic. 17 = 17 1 = 1 7 =, so the sequence is arithmetic, and the generating function is a n := 7 + (n 1). Let N denote the number that generates the last term of the sequence. This gives that a N = 117 = 7 + (N 1) 100 = N 10 = N 1 = N and we conclude there are 1 terms in the sequence. 3. Consider a sequence whose 3rd term is 13 and whose 7th term is 08. (a) Assume the sequence is arithmetic. Find the function a(n) that describes the sequence and find the 19th term of the sequence. Solution: We need to find the first term and the common difference = a 7 a 3 = (a 1 + 6d) (a 1 + d) = d and the common difference is 8.7. We can now find a 1 as 13 = a 3 = a 1 + (8.7) = a , and we conclude that a 1 = 8.. This gives that the generating function is a n := 8. + (n 1)(8.7). The 19 th term is a 19 = 793. (b) Assume the sequence is geometric. Find the function a(n) that describes the sequence and find the 19th term of the sequence. 08 Solution: We need to find the first term and common ratio. 13 = a 7 a 3 = a 1r 6 a 1 r = r and the common ratio is. We can now find a 1 as 13 = a 1 () = a 1, and we conclude that a 1 = 13. This gives that the generating functions is a n := 13 ()n 1. The 19 th term is a 19 = 13() 16

3 . Consider the infinite sequence 3k3, 9k, 7k, 81k6, (a) Determine the function that generates this geometric sequence. Solution: We know the sequence is geometric, so we must find the common ratio and first term. The common ratio r = 9k = 3k. The first term a 3k 3 1 = 3k3. Thus, the generating ( function is given by a n := a 1 (r) n 1 = 3k3 3k ) n 1 ( = 3 ) n k n+. (b) Determine the limit of this sequence. (Warning: Your answer must be dependent on the constant k.) Solution: From a theorem in class we have five scenarios; r > 1 and a 1 > 0, r > 1 and a 1 < 0, r 1, r = 1, and 1 r 1. Now, if r > 1, then we have the following: r > 1 3k > 1 a n =. If r 1, then we have the following: k > 3 a 1 > 0, r 1 3k 1 k 3 a n D.N.E. If 1 < r < 1, then we have the following: r = 1 3k = 1 k = 3 a n = a 1 = 9 If 1 < r < 1, then we have the following: 1 < r < 1 1 < 3k < 1 3 < k < 3

4 a n = 0 Putting it all together we have If k >, then lim a 3 n =. If k, then lim a 3 n D.N.E. If k =, then lim a 3 n = 9. If < k <, then lim a 3 3 n = 0.. Prove that the following sequences are neither arithmetic nor geometric. Then find the next terms each sequence assuming the simplest pattern continues. (a),, 7, 11, 16,... Solution: 7, so there is no common difference, and the sequence cannot be arithmetic. 7, so there is no common ratio, and the sequence cannot be geometric. The next four terms are, 9, 37, 6, as you add n + 1 to the nth term. (b) 1,, 0, 10, 0,... Solution: 1 0, so there is no common difference, and the sequence cannot be arithmetic. 0, so there is no common ratio, and the sequence cannot be geometric. 1 The next four terms are, 8, 8, 16, as you alternate dividing by and multiplying by. 6. For what value(s) of k will k 1, k + 3, 3k + 9 be a geometric sequence? Solution: We need k + 3 k 1 = 3k + 9 k + 3. We can either cross multiply and solve (k + 3) = (3k + 9)(k 1), or we can notice that 3k + 9 = 3(k + 3) and the previous equation becomes simplifies to k + 3 = 3(k 1) 6 = k 3 = k. If we cross multiply and solve the quadratic equation obtained (k + 3) = (3k + 9)(k 1), then we will get an extraneous solution of k = 3. This value for k will not work as it will make k + 3 = 0, and the sequence will be, 0, 0, 0,..., which is not a geometric series.

5 7. Evaluate each finite series. (a) Solution: This is an arithmetic series with common difference of. The generating function of the associated sequence is a n := 7 + (n 1). We must find the number of terms in order to apply the theorem from class for evaluating finite arithmetic series. To find N, where N is the number of terms, we solve 71 = 7 + (N 1)() for N and find that N = 17. The theorem gives us S = N(a 1+a N ) = 17(7+71) = 663. (b) The sum of where the associated sequence has 30 terms. Solution: This is an arithmetic series with common difference of 3. The generating function of the associated sequence is a n := 6 + (n 1)( 3). We must find the 30th term in order to apply the theorem from class for evaluating finite arithmetic series. To find a 30 we evaluate the generating function for the associated series with n = 30 and find that a 30 = 6 + (30 1)( 3) = 7. The theorem gives us S = 30(a 7 1+a 30 ) 30(6 = ) = 9 = How many terms of the sequence 13, 7, 1,, 11,... must be added to give a sum of 36? Solution: First note that the sequence is arithmetic with common difference d = 6. This means we can apply the formula from lecture and use that to solve for the number of terms. We know S = 36, a 1 = 13, and a N = 13 + (N 1)6. This gives N( 13 + ( 13 + (N 1)6)) 36 = 78 = N(6N 3) 78 = 6N 3N, and we must solve the quadratic equation 6N 3N 78 = 0 which factors into (N 1)(6N + ) = 0. However, only N = 1 makes sense as 6N + = 0 will yield neither a natural number nor a positive number. Thus, we conclude that we must add 1 terms of the sequence to give a sum of 36.

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