course Mth 151 ¶¹Ñ§©xÛéIƒÓñï¾»ùws¿áÎýÄ|²‹˜¹Èassignment #007
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07:21:40 Query 1.2.6 seq 2, 51, 220, 575, 1230, 2317 ... by successive differences
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RESPONSE --> 3992 is the answer. You break it down by the successive differences and it comes out to 24 which adds back up to 3992 confidence assessment: 3
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07:21:53 ** 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 --> Got it :-) self critique assessment: 3
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07:23:42 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 4^2+4=20 5^2-5=20 So yes it is true confidence assessment: 3
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07:24:08 1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3
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RESPONSE --> Ok. confidence assessment: 3
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07:27:14 ** 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 --> No number for this but it is not true 1+2+3=6 6^2=36 3^3=27 self critique assessment: 3
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07:29:04 1.2.36 1 st triangular # div by 3, remainder; then 2d etc. Pattern.
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RESPONSE --> No I'm not sure I see a pattern. Also I'm not sure what numbers to use. confidence assessment: 1
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07:29:40 ** 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 --> Ah ok I got that but I didn't think that was the answer wanted. self critique assessment: 3
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07:31:35 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 --> 408 confidence assessment:
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07:32:23 ** 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 --> Well I'm not sure if it saved my answer but the answer is 408 and you get it by plugging 12 in the n in the equation. self critique assessment: 3
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