Math 115 Spring 2011 Written Homework 5 Solutions


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1 . Evaluate each series. a) Math 5 Spring 0 Written Homework 5 Solutions Solution: We note that the associated sequence, 4, 7, 0,..., 55 appears to be an arithmetic sequence. If the sequence is arithmetic, then the common difference is d = 7 4 = 0 7 = 3. Then the candidate for the generating function is an) := 4 n )3. To prove that the associated sequence is arithmetic, we need to show that 55 is a term generated by the function an). That is, there is some N where an) = 55. an) = 55 4 N )3 = 55 3N = 55 3N = 54 N = 8 Since N = 8 is a natural number, the series is an arithmetic series. Using the summation formula for a finite arithmetic series, ) ) a a N 4 55 S = N = 8 = 959) = 53. b) The first and last terms of summation are 8 and, respectively and the common ratio 5 between each term is 4. Solution: Here we are given that the series ) is a geometric series. The generating function n for the associated sequence is bn) := 8. For a geometric series, we have two forms 4 of the summation formula, S = a N a r ) r N and S = a where N is the number of r
2 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 b = 8 and b N = 5. S = b N b r = 4 3 = b N)r b r = 5 ) 4 ) 8 4 ) = = ) = ) = = ) c) The sum of where the associated sequence has terms. Solution: We note that the associated sequence, 3, 6,, 4,... is a geometric sequence. r = 6 3 = 6 = 4 = The generating function for the sequence is cn) := 3) ) n. We are given that the series has terms. Here we use the other formulation for the sum of a finite geometric series. ) ) r N ) S = a = 3) = ). r ) d) The sum of 0, 5, 5, 5,... where the associated sequence has 30 terms. Solution: This is an arithmetic sequence: d = = 5 = 5 5 = 5. Then dn) := 0 n ) 5 ) is the generating function for the associated sequence. We are given that N = 30. Then ) ) )) a a 30 0 [0 30 ) 5 S = 30 = 30 ) ] = ) = 5 05 ) = 575.
3 . How many terms of the sequence 5,, 3,... must be added to give a sum of 400? Solution: We need to first determine if this sequence is arithmetic or geometric. Since 5) = 3 ) = 4, we assume that the sequence is arithmetic with a common difference d = 4. Then, the generating function of the sequence is an) := 5 n )4). The summation formula for the associated Nterm arithmetic series is ) ) a a N 5 5 N )4) S = N = N. We are given that the summation S = 400. We need to determine N. ) 5 5 N )4) 400 = N ) 0 4N 4 = N ) 4 4N = N = N 7 N) = 7N N 0 = N 7N 400 Factor the above quadratic equation or use the quadratic formula to solve for N. N = 7) ± 7) 4) 400). ) N = 7 ± 57 = 6 or 5. Now, both of these candidates for N can not be correct. Recall that N is a number of terms in a sequence / series. N must be a natural number. Thus, N = 6 is the only solution.
4 3. a) Use a series to find the sum of the first 00 odd, positive integers. Solution: We know from lecture that the sequence that generates the odd integer is oddn) := n ) = n. The series here is odd00) = [ 00)] = The summation formula for this arithmetic series is ) 399 S = 00 = b) Use a series to find the sum of all positive integers less than 00 that are multiples of 7. Solution: The series here is N where N is the largest natural number where 7N 00. Hence, N = 8. N 00 7 = We notice that the associated sequence is arithmetic: bn) := 7n. The summation formula for this arithmetic series is ) 7 78) S = 8 = 40) = 84.
5 4. How many terms of the sequence generated by the function a n := 43) n must be added to give a sum of 456? Solution: The associated series is clearly geometric. The summation formula for an N term geometric series is S = a r N r ) ) 3 N 456 = = 3N 78 = 3 N 79 = 3 N 79 = 3 N 3 6 = 3 N N = 6 There are 6 terms in the associated series. 5. If 0 a, 0 a, 0 a 3,..., 0 an is a geometric sequence, what can you determine about the sequence a, a, a 3,..., a n? Solution: We are given that 0 a, 0 a, 0 a 3,..., 0 an is a geometric sequence. Thus, r = 0a 0 a = 0a3 0 a = 0a4 0 a 3 =... = 0an 0 a n = 0 a a = 0 a 3 a = 0 a 4 a 3 =... = 0 an a n For the numbers to all be equal, the exponent on the base 0 must be the same. That is, there is some exponent p where p = a a = a 3 a = a 4 a 3 =... = a n a n. The equation a n a n = p for all n is the definition of an arithmetic sequence with common difference p. Hence, the sequence a, a, a 3,..., a n must be an arithmetic sequence.
6 9 6. Write π k as an expanded sum and compute the sum. k=3 9 Solution: π k = π 3 π 4 π 5 π 6 π 7 π 8 π 9. k=3 Recall that if a number x is negative then x = x. Additionally, recall that π Hence 6 < π < 7. Thus, π k = π k when k 6 and π k = π k), when k > 6. Then, 9 π k = π 3) π 4) π 5) π 6) [ π 7)] [ π 8)] [ π 9)] k=3 = π 3 π 4 π 5 π 6 π 7) π 8) π 9) = π 3 π 4 π 5 π 6 π 7 π 8 π 9 = π = π 6
7 7. Write each of the following using summation notation. a) a b ) a b ) 3 a 3 b 3 ) 4... a 0 b 0 ) 0 Solution: a i b i ) i. i= b) The sum of all three digit positive even integers. n= Solution: This is the series This associated sequence is arithmetic and is generated by the function bn) := 00 n ). Then the summation is N bn). We need to know how many terms are in the series in order to define the upperbound on the index in our summation notation. Note that we use bn) = 998 to determine how many terms are in the sequence. 998 = 00 N ) 898 = N ) 449 = N N = Then the series can be written in the form [00 n )]. n= Remark: An alternative method would result in the answer c) n=50 n. This is also correct. 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 = 6 = /3 = /9 /3 = 3. To be certain that this series is a geometric series, we need to show that the term 43 is a term in the sequence generated by cn) := 6 n. If it is, in this process we will 3)
8 determine how many numbers are in this sequence.) cn) = 43 6 ) N = 3 43 ) N = 3 79 N = 3) 3 N = 6 N = 7 ) 6 Thus, = 6 ) n 3 n=
9 4 8. If a b ab) = b= 5 ac 6), determine a. c=3 Solution: Here we expand and simplify. 4 a b ab) = b= 5 ac 6) c=3 a a) a 3 a3) a 4 a4) = a3 6) a4 6) a5 6) a a 3a 3a 4a 4a = 3a 6 4a 6 5a 6 9a 9a = a 8 9a a 8 = 0 3a 7a 6) = 0 3a 3)3a ) = 0 Here, the summations will be equal when a is either a = 3 or a =. 3
10 n 9. What is the sum of the series ) k, if n is odd? if n is even? Solution: Try some values for n. ) = ) = ) ) = = 0 3 ) = ) ) ) 3 = = 4 ) = ) ) ) 3 ) 4 = = 0 5 ) = ) ) ) 3 ) 4 ) 5 = = At this point, we recognize the pattern. If n is an even number, we can form n/ pairs of = 0 and the summation will always be 0. If n is an odd number, the pairs made by the first n terms will cancel and we will be left with a single. Hence, the summation equals  when n is odd.
11 0. a) Write 0 kk ) as an expanded sum and compute the sum. Solution: 0 kk ) = ) 3) 34) 45) 56) 67) 78) 89) 90) 0) = = 0 0 b) The summation k ) is an example of a telescoping sum. Expand and k compute this sum. What property of the summation makes this a telescoping sum? Solution: 0 k ) k = ) ) 3 3 ) 4 4 ) 5 5 ) 6 6 ) 7 7 ) 8 8 ) 9 9 ) 0 0 ) = = = 0 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 ). k Solution: N k ) k = ) ) 3 3 )... 4 N ) N = N or N N
12 d) Show that k k = kk ). Solution: e) Solution: Evaluate k k = k kk ) k kk ) = k k kk ) = kk ) 000 kk ) kk ) = k ) = k 00 =
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