course Mth 151 n}Zd݁assignment #007
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15:55:49 Query 1.2.6 seq 2, 51, 220, 575, 1230, 2317 ... by successive differences
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RESPONSE --> The next number is 3992 confidence assessment: 3
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15:55:56 ** If the sequence is 2, 57, 220, 575, 1230, 2317, ... then we have: 2, 57, 220, 575, 1230, 2317, # 3992 55, 163, 355, 655, 1087, # 1675 108, 192, 300, 432, # 588 84, 108, 132, # 156 24, 24, The final results, after the # signs, are obtained by adding the number in the row just below, in the following order: Line (4) becomes 132+24=156 Line (3) becomes 432+156=588 Line (2) becomes 1087+588=1675 Line (1) becomes 2317+1675=3992 The next term is 3992. **
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RESPONSE --> OK self critique assessment: 3
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15:56:41 1.2.18 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc.
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RESPONSE --> 4^2+4 = 5^2+5 This is a true statement confidence assessment: 3
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15:56:47 ** The next equation in the sequence would be 4^2 + 4 = 5^2 - 5 The verification is as follows: 4^2 + 4 = 5^2 - 5 simplifies to give you 16 + 4 = 25 - 5 or 20 = 20 **
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RESPONSE --> OK self critique assessment: 3
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15:58:10 1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3
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RESPONSE --> confidence assessment: 3
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ÚxRK_{ assignment #007 007. `Query 7 College Algebra 06-16-2007
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15:59:25 Query 1.2.6 seq 2, 51, 220, 575, 1230, 2317 ... by successive differences
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RESPONSE --> The next number is 3992 confidence assessment: 3
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15:59:35 ** If the sequence is 2, 57, 220, 575, 1230, 2317, ... then we have: 2, 57, 220, 575, 1230, 2317, # 3992 55, 163, 355, 655, 1087, # 1675 108, 192, 300, 432, # 588 84, 108, 132, # 156 24, 24, The final results, after the # signs, are obtained by adding the number in the row just below, in the following order: Line (4) becomes 132+24=156 Line (3) becomes 432+156=588 Line (2) becomes 1087+588=1675 Line (1) becomes 2317+1675=3992 The next term is 3992. **
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RESPONSE --> OK self critique assessment: 3
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16:00:12 1.2.18 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc.
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RESPONSE --> 4^2+4=5^2-5 20=20 This is a true statement confidence assessment: 3
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16:00:17 ** The next equation in the sequence would be 4^2 + 4 = 5^2 - 5 The verification is as follows: 4^2 + 4 = 5^2 - 5 simplifies to give you 16 + 4 = 25 - 5 or 20 = 20 **
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RESPONSE --> OK self critique assessment: 3
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16:00:49 1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3
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RESPONSE --> The square of the sum of the first n counting numbers is the same as the sum of the cubes of the first n counting numbers confidence assessment: 3
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16:00:54 ** the equation says that the square of the sum of the first n counting numbers is equal to the sum of their cubes **
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RESPONSE --> OK self critique assessment: 3
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16:08:55 1.2.36 1 st triangular # div by 3, remainder; then 2d etc. Pattern.
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RESPONSE --> 1/3= .3 remainder is .1 3/3=1 no remainder 6/3=2 no remainder 10/3=3.3 remainder is .1 15/3=5 no remainder 21/3= 7 no remainder 28/3= 9.3 remainder is 1 The pattern is .1, 0, 0, .1, 0, 0, .1, ....... confidence assessment: 2
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16:10:38 ** The triangular numbers are 1, 3, 6, 10, 15, 21, . . . . We divide these by 3 and get the sequence of remainders. When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0. It turns out that the sequence continues as a string of 1,0,0 's. At this point that is an inductive pattern, but remmeber that the sequence of triangular numbers continues by adding successively larger and larger numbers to the members of the sequence. Since the number added always increases by 1, and since every third number added is a multiple of 3, is isn't too difficult to see how the sequence of remainders comes about and to see why it continues as it does. COMMON ERROR: .3333333,1,2,3.3333333,etc. INSTRUCTOR CORRECTION: You need the remainders, not the decimal equivalents. When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's. COMMON ERROR: 1/3, 1, 2, 3 1/3 CORRECTION: These are the quotients. You need the remainders. If you get 1/3 that means the remainder is 1; same if you get 3 1/3. If you just getting whole number (like 1 or 2 in your calculations) the remainder is 0. In other words, when you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. The remainders form a sequence 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's. **
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RESPONSE --> I took it out to a decimal instead of using a whole number self critique assessment: 2
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16:12:05 1.2.48 use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.
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RESPONSE --> To fine the nth octagonal number, you find n(6n-4) and divide that number by 2 12(6*12-4)= 816 816/2= 408 confidence assessment: 3
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16:12:12 ** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows: Triangular numbers: n / 2 * [ n + 1 ] note that this is the same as Gauss' formula Square numbers: n / 2 * [ 2n + 0 ] or just n^2 Pentagonal #'s: n / 2 * [ 3n - 1 ] Hexagonal #'s: n / 2 * [ 4n - 2 ] Heptagonal #'s: n / 2 * [ 5n - 3 ] Octagonal #'s: n / 2 * [ 6n - 4 ] The coefficient of n in the bracketed term starts with 1 and increases by 1 each time, and the +1 in the first bracketed term decreases by 1 each time. You will need to know these formulas for the test. The last formula is for octagonal numbers. To get n = 12 octangonal number use n/2 * [ 6n - 4 ] to get 12 / 2 * [ 6 * 12 - 4 ] = 6 * 68 = 408. **
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RESPONSE --> OK self critique assessment: 3
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