#$&* course Mth 151 2/1 8 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `a** Using sequences of differences we obtain: 2, 57, 220, 575, 1230, 2317, # 3992 55, 163, 355, 655, 1087, # 1675 108, 192, 300, 432, # 588 84, 108, 132, # 156 24, 24, The final results, after the # signs, are obtained by adding the number in the row just below, in the following order: Line (4) becomes 132+24=156 Line (3) becomes 432+156=588 Line (2) becomes 1087+588=1675 Line (1) becomes 2317+1675=3992 The next term is 3992. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 1.2.30 (previously 1.2.18) 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1^2 + 1 = 2^2 - 2 =2 2^2 + 2 = 3^2 - 3 =6 3^2 + 3 = 4^2 - 4 =12 4^2 + 4 = 5^2 - 5 =20 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I might should have done a little better of a job verifying. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q1.2.42 (previously 1.2.30) state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The addition of all the counting numbers squared is equal to the addition of cubed counting numbers. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** the equation says that the square of the sum of the first n counting numbers is equal to the sum of their cubes ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Ummm…. The answer sounds better than mine. ------------------------------------------------ Self-critique Rating: 2
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Given Solution: `a** 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. ** Self-critique (if necessary) : Oh, I seem to have misunderstood the problem. Well, at least what I discovered was pretty neat. ------------------------------------------------ Self-critique Rating:0
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Given Solution: `a** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows (you should know this pattern; you might need one or more of these formulas on the test): 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. ** Ah! I see. " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q 1.2. 58 (previously 1.2.48) use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Psubn= n(6n-1)/2 Psub12= 12(6*12 - 1 )/2 Psub12= 12(71)/2 = 852/2 426 426 is the 12th octagonal number. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows (you should know this pattern; you might need one or more of these formulas on the test): 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. ** Ah! I see. " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `q 1.2. 58 (previously 1.2.48) use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Psubn= n(6n-1)/2 Psub12= 12(6*12 - 1 )/2 Psub12= 12(71)/2 = 852/2 426 426 is the 12th octagonal number. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows (you should know this pattern; you might need one or more of these formulas on the test): 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. ** Ah! I see. " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!