Math 115 Spring 2011 Homework Solutions: Series and Sequences - Prof. Jennifer R. Mcneilly, Study notes of Mathematics

Download Math 115 Spring 2011 Homework Solutions: Series and Sequences - Prof. Jennifer R. Mcneilly and more Study notes Mathematics in PDF only on Docsity! Math 115 Spring 2011 Written Homework 5 Solutions 1. Evaluate each series. (a) 4 + 7 + 10 + ...+ 55 Solution: We note that the associated sequence, 4, 7, 10, . . . , 55 appears to be an arithmetic sequence. If the sequence is arithmetic, then the common difference is d = 7−4 = 10−7 = 3. Then the candidate for the generating function is a(n) := 4 + (n− 1)3. To prove that the associated sequence is arithmetic, we need to show that 55 is a term generated by the function a(n). That is, there is some N where a(N) = 55. a(N) = 55 4 + (N − 1)3 = 55 3N + 1 = 55 3N = 54 N = 18 Since N = 18 is a natural number, the series is an arithmetic series. Using the summation formula for a finite arithmetic series, S = N ( a1 + aN 2 ) = 18 ( 4 + 55 2 ) = 9(59) = 531. (b) The first and last terms of summation are 8 and 1 512 , respectively and the common ratio between each term is 1 4 . Solution: Here we are given that the series is a geometric series. The generating function for the associated sequence is b(n) := 8 ( 1 4 )n−1 . For a geometric series, we have two forms of the summation formula, S = aN+1 − a1 r − 1 and S = a1 ( rN − 1 r − 1 ) where N is the number of terms in the series. While we could find N using the generating function, since we know the first and last terms in the series, we don’t need to. Here b1 = 8 and bN = 1 512 . S = bN+1 − b1 r − 1 = (bN)r − b1 r − 1 = ( 1 512 )(1 4 )− 8 (1 4 )− 1 = 1 2048 − 8 −3 4 = −4 3 ( 1 2048 − 8 ) = −4 3 ( 1 2048 − 16384 2048 ) = −4 3 ( −16383 2048 ) = 65532 6144 = 5461 512 (c) The sum of −3 + 6− 12 + 24− ... where the associated sequence has 21 terms. Solution: We note that the associated sequence, −3, 6,−12, 24, . . . is a geometric sequence. r = 6 −3 = −12 6 = 24 −12 = −2 The generating function for the sequence is c(n) := (−3)(−2)n−1. We are given that the series has 21 terms. Here we use the other formulation for the sum of a finite geometric series. S = a1 ( rN − 1 r − 1 ) = (−3) ( (−2)21 − 1 (−2)− 1 ) = (−2)21 − 1. (d) The sum of 10, 15 2 , 5, 5 2 , ... where the associated sequence has 30 terms. Solution: This is an arithmetic sequence: d = 15 2 − 10 = 5− 15 2 = 5 2 − 5 = −5 2 . Then d(n) := 10 + (n− 1) ( −5 2 ) is the generating function for the associated sequence. We are given that N = 30. Then S = 30 ( a1 + a30 2 ) = 30 ( 10 + [10 + (30− 1) ( −5 2 ) ] 2 ) = 30 ( 20 + 29 ( −5 2 ) 2 ) 15 ( 20− 145 2 ) = 15 ( −105 2 ) = −1575 2 . 4. How many terms of the sequence generated by the function an := 4(3) n−1 must be added to give a sum of 1456? Solution: The associated series is clearly geometric. The summation formula for an N term geometric series is S = a1 ( 1− rN 1− r ) 1456 = 4 ( 1− 3N 1− 3 ) 364 = 1− 3N −2 −728 = 1− 3N −729 = −3N 729 = 3N 36 = 3N N = 6 There are 6 terms in the associated series. 5. If 10a1 , 10a2 , 10a3 , ..., 10an is a geometric sequence, what can you determine about the sequence a1, a2, a3, ..., an? Solution: We are given that 10a1 , 10a2 , 10a3 , ..., 10an is a geometric sequence. Thus, r = 10a2 10a1 = 10a3 10a2 = 10a4 10a3 = . . . = 10an 10an−1 = 10a2−a1 = 10a3−a2 = 10a4−a3 = . . . = 10an−an−1 For the numbers to all be equal, the exponent on the base 10 must be the same. That is, there is some exponent p where p = a2 − a1 = a3 − a2 = a4 − a3 = . . . = an − an−1. The equation an−an−1 = p for all n is the definition of an arithmetic sequence with common difference p. Hence, the sequence a1, a2, a3, ..., an must be an arithmetic sequence. 6. Write 9∑ k=3 |2π − k| as an expanded sum and compute the sum. Solution: 9∑ k=3 |2π−k| = |2π−3|+ |2π−4|+ |2π−5|+ |2π−6|+ |2π−7|+ |2π−8|+ |2π−9|. Recall that if a number x is negative then |x| = −x. Additionally, recall that π ≈ 3.14159 . . .. Hence 6 < 2π < 7. Thus, |2π − k| = 2π − k when k ≤ 6 and |2π − k| = −(2π − k), when k > 6. Then, 9∑ k=3 |2π − k| = (2π − 3) + (2π − 4) + (2π − 5) + (2π − 6) + [−(2π − 7)] + [−(2π − 8)] + [−(2π − 9)] = 2π − 3 + 2π − 4 + 2π − 5 + 2π − 6− (2π − 7)− (2π − 8)− (2π − 9) = 2π − 3 + 2π − 4 + 2π − 5 + 2π − 6− 2π + 7− 2π + 8− 2π + 9 = 2π − 3− 4− 5− 6 + 7 + 8 + 9 = 2π + 6 7. Write each of the following using summation notation. (a) a1(b1) 2 + a2(b2) 3 + a3(b3) 4 + ...+ a10(b10) 11 Solution: 10∑ i=1 ai(bi) i+1. (b) The sum of all three digit positive even integers. Solution: This is the series 100 + 102 + 104 + . . .+ 996 + 998. This associated sequence is arithmetic and is generated by the function b(n) := 100 + (n− 1)2. Then the summation is N∑ n=1 b(n). We need to know how many terms are in the series in order to define the upper- bound on the index in our summation notation. Note that we use b(N) = 998 to determine how many terms are in the sequence. 998 = 100 + (N − 1)2 898 = (N − 1)2 449 = N − 1 N = 450 Then the series can be written in the form 450∑ n=1 [100 + (n− 1)2]. Remark: An alternative method would result in the answer 499∑ n=50 2n. This is also correct. (c) 6− 2 + 2 3 − 2 9 + ...+ 2 243 Solution: Again, in order to write this as a summation, we need a generating function for the associate sequence. We are lead to believe that the associate sequence is geometric because r = −2 6 = 2/3 −2 = −2/9 2/3 = −1 3 . To be certain that this series is a geometric series, we need to show that the term 2 243 is a term in the sequence generated by c(n) := 6 ( −1 3 )n−1 . (If it is, in this process we will 9. What is the sum of the series n∑ k=1 (−1)k, if n is odd? if n is even? Solution: Try some values for n. 1∑ k=1 (−1) = −1 2∑ k=1 (−1) = (−1) + (−1)2 = −1 + 1 = 0 3∑ k=1 (−1) = (−1) + (−1)2 + (−1)3 = −1 + 1− 1 = −1 4∑ k=1 (−1) = (−1) + (−1)2 + (−1)3 + (−1)4 = −1 + 1− 1 + 1 = 0 5∑ k=1 (−1) = (−1) + (−1)2 + (−1)3 + (−1)4 + (1)5 = −1 + 1− 1 + 1− 1 = −1 At this point, we recognize the pattern. If n is an even number, we can form n/2 pairs of −1 + 1 = 0 and the summation will always be 0. If n is an odd number, the pairs made by the first n− 1 terms will cancel and we will be left with a single −1. Hence, the summation equals -1 when n is odd. 10. (a) Write 10∑ k=1 1 k(k + 1) as an expanded sum and compute the sum. Solution: 10∑ k=1 1 k(k + 1) = 1 1(2) + 1 2(3) + 1 3(4) + 1 4(5) + 1 5(6) + 1 6(7) + 1 7(8) + 1 8(9) + 1 9(10) + 1 10(11) = 1 2 + 1 6 + 1 12 + 1 25 + 1 30 + 1 42 + 1 56 + 1 72 + 1 90 + 1 110 = 10 11 (b) The summation 10∑ k=1 ( 1 k − 1 k + 1 ) is an example of a telescoping sum. Expand and compute this sum. What property of the summation makes this a “telescoping sum”? Solution: 10∑ k=1 ( 1 k − 1 k + 1 ) = ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + ( 1 4 − 1 5 ) + ( 1 5 − 1 6 ) + ( 1 6 − 1 7 ) + ( 1 7 − 1 8 ) + ( 1 8 − 1 9 ) + ( 1 9 − 1 10 ) + ( 1 10 − 1 11 ) = 1 1 − 1 2 + 1 2 − 1 3 + 1 3 − 1 4 + 1 4 − 1 5 + 1 5 − 1 6 + 1 6 − 1 7 + 1 7 − 1 8 + 1 8 − 1 9 + 1 9 − 1 10 + 1 10 − 1 11 = 1− 1 11 = 10 11 The property that makes this a “telescoping sum” is the fact that it collapses down to a much smaller sum (just as a telescope can expand and collapse). (c) Let N be a large positive number. Evaluate N∑ k=1 ( 1 k − 1 k + 1 ) . Solution: N∑ k=1 ( 1 k − 1 k + 1 ) = ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + . . .+ ( 1 N − 1 N + 1 ) = 1− 1 N + 1 or N 1 +N (d) Show that 1 k − 1 k + 1 = 1 k(k + 1) . Solution: 1 k − 1 k + 1 = k + 1 k(k + 1) − k k(k + 1) = k + 1− k k(k + 1) = 1 k(k + 1) (e) Evaluate 1000∑ k=1 1 k(k + 1) Solution: 1000∑ k=1 1 k(k + 1) = 1000∑ k=1 ( 1 k − 1 k + 1 ) = 1− 1 1001 = 1000 1001