course mth164 end program˙ͨՙYӧCܛ{BV
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11:41:59 `q001. What is the pattern of the sequence 2 6 10 14 18 ... ? If this pattern holds what is the value of the 100th member of this sequence? In terms of the symbol n, what is the value of the nth number in this sequence?
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RESPONSE --> the pattern is +4 and the 100th term would be 396 dont know what nth number means? confidence assessment: 2
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11:43:26 The pattern is fairly easy to see. Each member of the sequence is 4 greater than the preceding member. To get to the 2d member of the sequence we can start at 2 and make a 'jump' of 4 units each, ending up at 6. To get to the 3d member of the sequence we can start at 2 and make 2 'jumps' of 4 units each. The number of 'jumps' necessary to reach the numbers in the second and third positions of the sequence will be 1 and 2, respectively. The number of 'jumps' therefore appears to the 1 less than the position of the number. It should be clear that this pattern will continue, with each succeeding position of the sequence corresponding to an additional 'jump' of 4 units. It follows that the 100th member of the sequence can be found by starting at 2 and making 99 'jumps' of 4 each. 99 'jumps' of 4 each will make a net 'jump' of 4 * 99 = 396 units. Starting from 2 this will bring us to 396 * 2 = 398. The 100th member of the sequence will therefore be 2 + 99 * 4 = 398.
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RESPONSE --> ok forgot to add the 2 which was the starting number still dont understand what the nth number is
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11:47:14 `q002. What is the pattern of the sequence 1 2 3 4 ... ? What is the value of the 100th member of this sequence? What is the sum of the first 100 numbers in this sequence? What is the value of the nth member of this sequence?
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RESPONSE --> pattern is +1 100 5050 confidence assessment: 2
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11:48:29 The pattern of this sequence is as obvious as it seems. We are adding 1 with each additional position. The 1st member is 1, the 2d member is 2, the 3d member is 3, etc.. The 100th member is 100. To get the sum of the first 100 members of the sequence we can use the pattern illustrated in Figure 102, grouping 1 with 100 to get 1 + 100 = 101, then 2 with 99 to get 2 + 99 = 101, then 3 with 98 to get 3 + 98 = 101, etc.. It should be clear that following this pattern we will obtain 1 such pair for every 2 numbers, for a total of 100 / 2 = 50 pairs. Each pair will total 101 so the sum of all the numbers is 50 * 101 = 5050.
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RESPONSE --> ok self critique assessment: 2
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11:50:49 `q003. If we want to find the sum of the first n members of the sequence 1 2 3 4 ... , we can as in the solution to the preceding question pair the numbers in such a way that each pair adds up to the same thing. How many pairs will we get from the first n members of the sequence? What is the first of the n members and what is the last? What is their sum? What is the second of the n member and the second to the last? What is their sum? What pattern do you see here and how can it help you find the sum of the first n members? What is your expression for the sum of the first n members of the sequence?
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RESPONSE --> ok dont fully understand quiestion confidence assessment: 2
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11:52:31 To get the sum of the first n members of the sequence we can use the pattern illustrated in Figure 107, grouping 1 with n to get 1 + n = n + 1, then 2 with n - 1 to get 2 + (n-1) = n + 1, then 3 with n - 2 to get 3 + (n - 2) = n + 1, etc.. It should be clear that following this pattern we will obtain 1 such pair for every 2 numbers, for a total of n / 2 pairs. Each pair will total n + 1 so the sum of all the numbers is n/2 * (n + 1) = n ( n + 1) / 2.
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RESPONSE --> ok i see what the problem is explaining self critique assessment: 2
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11:56:52 `q004. Adapt the reasoning process used in the preceding two questions to find the sum of the first 100 members of the sequence 2 6 10 14 18 ... . Find the expression for the nth member and determine the sum of the first n members of this sequence.
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RESPONSE --> 20000 confidence assessment: 2
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11:57:06 In the first problem we saw that the 100th member of the sequence 2 6 10 14 ... is 2 + 99 * 4 = 398. The sum of the first and last numbers is therefore 2 + 389 = 400, while the sum of the second and next-to-last numbers is 6 + 394 = 400. It should be clear that we can continue pairing the numbers in this manner, obtaining the sum 400 for each pair. We have 100 numbers so we have 100 / 2 = 50 pairs. The total of 50 pairs each adding up to 400 is 50 * 400 = 20,000. The first 100 numbers in this sequence therefore total 20,000.
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RESPONSE --> ok self critique assessment: 2
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12:00:43 `q005. What is the pattern of the sequence 1 2 4 8 16 32 . . . ? What is the 30th member of this sequence? What is the expression for the nth member of the sequence?
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RESPONSE --> x2 1073741824 confidence assessment: 2
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12:02:17 Each member of this sequence is double the preceding member. The second element is 2 * 1, the third is 4 = 2 * 2 = 2 * 2, the fourth is 8 = 2 * 4 = 2 * 2 * 2, etc.. Another way of saying this is to say that the second element is 2^1, the third is 2^2, the fourth is 2^3, etc.. The power of 2 is always 1 less than the position in the sequence. The 30th member will therefore be 2^(30-1) = 2^29 = 536,870,912.
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RESPONSE --> used 30 instead of 29 as a power i now see that you use one less because you start at the 2nd number self critique assessment: 2
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12:11:02 `q006. What is the pattern of the sequence 6 12 24 48 96 192 . . . ? What is the 30th member of this sequence? What is the expression for the nth member of the sequence?
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RESPONSE --> 2.21^23 confidence assessment: 2
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12:12:30 Each member of this sequence is double the preceding member. The second element is 12 = 2 * 6, the third is 24 = 2 * 2 * 6 = 2^2 * 6, the fourth is 48 = 2 * 2 * 2 * 6 = 2^3 * 6, etc.. Again the power of 2 is always 1 less than the position in the sequence, and is multiplied by the first number in the sequence (i.e., by 6 in this example). The 30th member will therefore be 6 * 2^(30-1) = 6 * 2^29 = 3,221,225,472.
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RESPONSE --> ok i see what the equation is it is the first term * 2^29 i thought the 2 was the first or second term in the string now i see self critique assessment: 2
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12:13:59 `q007. What is the pattern of the sequence 4 2 1 1/2 1/4 1/8 . . . ? What is the 30th member of this sequence? What is the expression for the nth member of the sequence?
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RESPONSE --> /2 30th term is 7.45^-9 confidence assessment: 2
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12:14:12 Each member of this sequence is half the preceding member. The second element is 2 = 1/2 * 4, the third is 1 = 1/2 * 1/2 * 4 = 1/2^2 * 4, the fourth is 1/2 = 1/2 * 1/2 * 1/2 * 4 = 1/2^3 * 4, etc.. The power of 1/2 is always 1 less than the position in the sequence, and is multiplied by the first number in the sequence (i.e., by 4 in this example). The 30th member will therefore be 4 * 1/2^(30-1) = 4 * 1/2^29 = 4 * 1/536870912 = 1/134217728 (this is exact, and is approximately equal to .000000007450580596, or equivalently to 7.450580596 * 10^-9).
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RESPONSE --> ok self critique assessment: 3
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12:21:39 `q008. The second partial sum of the sequence 4 2 1 1/2 1/4 1/8 . . . is 4 + 2 = 6. The third partial sum is 4 + 2 + 1 = 7. What are the fourth, fifth, sixth, seventh and eighth partial sums? Do the partial sums of this sequence have an upper bound--i.e., is there a number the partial sums can never exceed?
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RESPONSE --> 4+2+1+1/2=7.5 4+2+1+.5+.25=7.75 4+2+1+.5+.25+.125=7.875 and so on ....... there is no number of partial sums that can be exceeded. It will continue to a point to where the sums will be increasing but very very minimally confidence assessment: 2
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12:21:49 The fourth partial sum is 4 + 2 + 1 + 1/2 = 7 1/2. The fifth partial sum is 4 + 2 + 1 + 1/2 + 1/4 = 7 3/4. The sixth partial sum is 4 + 2 + 1 + 1/2 + 1/4 + 1/8 = 7 7/8. The seventh partial sum is 4 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 = 7 15/16. The eighth partial sum is 4 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 7 31/32.
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RESPONSE --> ok self critique assessment: 2
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12:24:02 `q009. The partial sums of the sequence 4 2 1 1/2 1/4 1/8 . . . are 4, 6, 7, 7 1/2, 7 3/4, 7 7/8, 7 15/16, ... . What is the pattern to these partial sums? Will the partial sum ever exceed 8? Is there any limit to how close the partial sums can get to 8?
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RESPONSE --> it cant pass eight but it will take a very very long time to get close due to the sums added every time are being cut in half. every sum it will get closer but never reach it confidence assessment: 2
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12:24:23 The partial sums can be compared with 8: 7 1/2 is 1/2 less than 8 7 3/4 is 1/4 less than 7 7 7/8 is 1/8 less than 8 7 15/16 is 1/16 less than 8 7 31/32 is 1/32 less than 8. Each partial sum is twice as close to 8 as the last number added. Since the last number added will always be positive but will eventually be as small as we might wish, if this pattern continues (as it does) we expect that the partial sums will approach as close as we wish to 8 without ever reaching 8.
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RESPONSE --> ok self critique assessment: 2
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12:28:00 `q010. If a sequence can be written in the form a * r^0, a * r^1, a * r^2, a * r^3, ..., a * r^n, ... that sequence is said to be a geometric sequence with common ratio r. How can the sequence 4, 2, 1, 1/2, 1/4, 1/8, . . . be expressed in the given form? Give the values of r and a that put the sequence into this form.
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RESPONSE --> when the values are placed in the form as given the a from the preceeding number fills in for the next in the sequence and you get the number values shown on the list below the r and a form confidence assessment: 2
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12:28:14 The sequence 4, 2, 1, 1/2, 1/4, 1/8, . . . be expressed in the given form by first factoring out the 4 to obtain the form 4 ( 1, 1/2, 1/4, 1/8, . . . ). The common ratio of the sequence is found by dividing each member of the sequence by its predecessor: r = 2/4 = 1/2, or r = 1/2, or r = 1/2 / (1/4) = 1/2. We always get the same result so r = 1/2. The sequence 1, 1/2, 1/4, 1/8, . . . is expressed in terms of powers of r as 1, 1/2, 1/4, 1/8, . . . = (1/2)^0, (1/2)^1, (1/2)^2, (1/2)^3, . . . = r^0, r^1, r^2, r^3, . . . Thus we say that sequence 4, 2, 1, 1/2, 1/4, 1/8, . . . is expressed in standard form a * r^0, a * r^1, a * r^2, a * r^3, ..., a * r^n, . . . as 4 ( (1/2)^0, (1/2)^1, (1/2)^2, (1/2)^3, . . . ).
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RESPONSE --> ok self critique assessment: 2
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12:28:37 `q011. The sequence r^0, r^1, r^2, ... , r^n, ... has the property that its (n+1)th partial sum, which is 1 + r + r^2 + r^3 + ... + r^n, is equal to Sn = ( 1 - r^n ) / ( 1 - r). What therefore is the value of the 11th partial sum of the sequence 1, 1/2, 1/4, 1/8, ..., 1/n, ... ?
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RESPONSE --> ok confidence assessment: 2
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12:31:15 The 11th partial sum is the n = 11-1 = 10 value of the expression Sn = (1 - r^n ) / ( 1 - r ) for the sequence 1, 1/2, 1/4, 1/8, . . ., 1/n, . . . with r = 1/2. We get S10 = (1 - (1/2)^10) / ( 1 - 1/2) = (1 - 1/1024) / (1 - 1/2) = (1023 / 1024) / (1/2) = 2 * (1023 / 1024) = 2046 / 1024 = 1 511/512.
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RESPONSE --> ok self critique assessment: 2
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12:34:17 `q012. What is the value of the 10th partial sum of the sequence 4, 2, 1, 1/2, 1/4, 1/8, ... ? What is the value of the nth partial sum of this sequence?
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RESPONSE --> 7.9296875 = partial sum of 10th confidence assessment: 2
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12:35:07 As seen previously we can write this sequence as 4 (1 + 1/2 + 1/4 + 1/8 + . . . + (1/2)^n + . . .). The 11th partial sum of 1 + 1/2 + 1/4 + 1/8 + . . . + (1/2)^n + . . . is the n = 11-1 = 10 value of the expression Sn = (1 - r^n ) / ( 1 - r ) with r = 1/2. We get S10 = (1 - (1/2)^10) / ( 1 - 1/2) = (1 - 1/1024) / (1 - 1/2) = (1023 / 1024) / (1/2) = 2 * (1023 / 1024) = 2046 / 1024. The 10th partial sum of the original sequence is thus 4 (1 + 1/2 + 1/4 + 1/8 + . . . + (1/2)^n + . . .) = 4 * 2046 / 1024 = 2046 / 256 = 7 254/256 = 7 127/128.
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RESPONSE --> reason for differance is the calculator rounded the fractions instead of using the fraction form self critique assessment: 2
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12:40:08 `q013. A sequence is defined by the expression a(n) = 4 n - 3. What are the values of a(1), a(2), a(3) and a(4)? What are the first four partial sums of the sequence?
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RESPONSE --> 1=1 1+5=6 1+5+9=15 1+5+9+13=28 confidence assessment: 2
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12:40:21 To get a(1) we substitute n = 1 into the expression a(n) = 4 n - 3, obtaining a(1) = 4 * 1 - 3 = 1. Similarly we get a(2) = 4 * 2 - 3 = 5, a(3) = 4 * 3 - 3 = 9, a(4) = 4 * 4 - 3 = 13. The partial sums are S1 = 1, S2 = 1 + 5 = 6, S3 = 1 + 5 + 9 = 15, S4 = 1 + 5 + 9 + 13 = 28.
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RESPONSE --> ok correct self critique assessment: 2
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12:42:25 `q014. A sequence is defined by the expression a(n) = 4 * 3^(n-1). What are the values of a(1), a(2), a(3) and a(4)? What are the first four partial sums of the sequence?
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RESPONSE --> 4,12,36,108 4+12=16 4+12+36=52 4+12+36+108=160 confidence assessment: 2
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12:42:34 To get a(1) we substitute n = 1 into the expression a(n) = 4 * 3^(n-1), obtaining a(1) = 4 * 3^(1-1) = 4 * 3^0 = 4 * 1 = 4. Similarly we get a(2) = 4 * 3^(2-1) = 4 * 3 = 12, a(3) = 4 * 3^(3-1) = 4 * 3^2 = 36, a(4) = 4 * 3^(4-1)= 4 * 3^3 = 4 = 108. The partial sums are S1 = 4, S2 = 4 + 12 = 16, S3 = 4 + 12 + 36 = 52, S4 = 4 + 12 + 36 + 108 = 160.
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RESPONSE --> ok self critique assessment: 2
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12:45:00 `q015. A sequence is defined by the expression a(n) = 4 * (-1/2)^(n-1). What are the values of a(1), a(2), a(3) and a(4)? What are the first four partial sums of the sequence?
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RESPONSE --> 4,-2,1,-.5 4+-2=2 4+-2+1=3 4+-2+1+-.5=2.5 confidence assessment: 2
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12:46:22 To get a(1) we substitute n = 1 into the expression a(n) = 4 * (-1/2)^(n-1), obtaining a(1) = 4 * (-1/2)^(1-1) = 4 * (-1/2)^0 = 4 * 1 = 4. Similarly we get a(2) = 4 * (-1/2)^(2-1) = 4 * (-1/2) = -2, a(3) = 4 * (-1/2)^(3-1) = 4 * (-1/2)^2 = 1, a(4) = 4 * (-1/2)^(4-1)= 4 * (-1/2)^3 = -1/2.. The partial sums are S1 = 4, S2 = 4 + (-2) = 2, S3 = 4 + (-2) + 1 = 3, S4 = 4 + (-2) + 1 + (-1/2) = 2 1/2.
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RESPONSE --> the preceeding equation given had a negative for the 1/2 i thought was this correct but anyway i see why i missed this one self critique assessment: 2
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