#$&* course Mth 151 1st time submitting this assignemtn03/21/2011 0619PM If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** Using sequences of differences we obtain: 2, 57, 220, 575, 1230, 2317, # 3992 55, 163, 355, 655, 1087, # 1675 108, 192, 300, 432, # 588 84, 108, 132, # 156 24, 24, The final results, after the # signs, are obtained by adding the number in the row just below, in the following order: Line (4) becomes 132+24=156 Line (3) becomes 432+156=588 Line (2) becomes 1087+588=1675 Line (1) becomes 2317+1675=3992 The next term is 3992. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n?a ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q1.2.18 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1^2 + 1 = 2^2 - 2 does work, 2^2 + 2 = 3^2 - 3does work, 3^2 + 3 = 4^2-4 does work, The next sequence will be 4^2 + 4 = 5^5 - 5 = 20 and this is also correct confidence rating #$&*: 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): N/A ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sum up the left side of the equation and take the square of it. For the right side, take the cubed of each individual number. They will both equal each other on both sides confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** the equation says that the square of the sum of the first n counting numbers is equal to the sum of their cubes ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): N/A ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q1.2.36 1 st triangular # div by 3, remainder; then 2d etc. Pattern. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/3= .33, 3/3= 1, 6/3= 2, 10/3= 3.33, 15/3= 15, Yes, there is a patter. It seems as though every 1st number is a remainder of .33 and inbetween that ones, whole numbers take the place for the 2nd and 3rd number. confidence rating #$&*: 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): I made the common error. After seeing the common error, this makes much more sense. I did not know the difference between quotients/remainders. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q1.2.48 use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: do not have a formula for octagonal figures so I am using the pentagonal equation: Pn = ( n ( 3 (n) - 1 )) / 2. This gives me: P(12) = ( 12 ( 3 (12) - 1 )) / 2 = 210 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows (you should know this pattern; you might need one or more of these formulas on the test): Triangular numbers: n / 2 * [ n + 1 ] note that this is the same as Gauss' formula Square numbers: n / 2 * [ 2n + 0 ] or just n^2 Pentagonal #'s: n / 2 * [ 3n - 1 ] Hexagonal #'s: n / 2 * [ 4n - 2 ] Heptagonal #'s: n / 2 * [ 5n - 3 ] Octagonal #'s: n / 2 * [ 6n - 4 ] The coefficient of n in the bracketed term starts with 1 and increases by 1 each time, and the +1 in the first bracketed term decreases by 1 each time. You will need to know these formulas for the test. The last formula is for octagonal numbers. To get n = 12 octangonal number use n/2 * [ 6n - 4 ] to get 12 / 2 * [ 6 * 12 - 4 ] = 6 * 68 = 408. ** With the given equation: n / 2 * [ 6 (n) - 4 ], this gives me: 12 / 2 * [ 6 (12) - 4 ] = 6 * 68 = 408 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q1.2.48 use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: do not have a formula for octagonal figures so I am using the pentagonal equation: Pn = ( n ( 3 (n) - 1 )) / 2. This gives me: P(12) = ( 12 ( 3 (12) - 1 )) / 2 = 210 confidence rating #$&*: 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. ** With the given equation: n / 2 * [ 6 (n) - 4 ], this gives me: 12 / 2 * [ 6 (12) - 4 ] = 6 * 68 = 408 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!