#$&* course MTH 151 007. Triangular, Square, Pentagonal Numbers
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Given Solution: The original triangle had the three points A, B and C. When you extended the two sides you marked the new endpoints, then you marked the point in the middle of the third side. So you've got 6 points marked. The construction of these numbers is shown in the figure below. We begin with a single dot: We label this point A and construct a triangle containing this point as a vertex. We place similar dots at the vertices of this triangle. We now 'scale up' the triangle by doubling the lengths of its sides: We divide this triangle into triangles of the original size, and place dots at each of these vertices. The first figure has a single 'dot', the second has 3 'dots', and the third has 6 'dots'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Extend the two sides that meet at A by distances equal to the original lengths AC and AB and mark the endpoints of the newly extended segments. Each of the newly extended sides will have 4 marked points. Now connect the new endpoints to form a new right triangle. Mark points along the new side at the same intervals that occur on the other two sides. How many marked points are on your new triangle, and how many in the whole figure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 2 marked points on the new triangle There are 10 marked points in the whole figure confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You added the two new endpoints when you extended the sides. You then should have marked two new points on the new third side, so that each side contains 4 points including its endpoints. Your figure will now contain 10 marked points. The construction is shown below. First we extend the two sides by a length equal to that of the original triangle: Next we join the 'free' endpoints of those new sides to form a triangle. Now we place points along the new side and join them to complete the 'small' triangles within our new figure: We have added four new dots. The figure below depicts only the 'dots', without the triangles: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Continue the process for another step-extend each side by a distance equal to the original point-to-point distance. How many points do you have in the new triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 15 points confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You will add an endpoint to each newly extended side, so each of the new sides will contain 5 points. You will then have to add 3 equally spaced points to the new side, giving you a total of 13 points on the new triangle. In addition there are two marked points inside the triangle, for a total of 15 points. Click on 'Next Picture' to see the construction. The line segments along two sides of the triangle have again been extended and points marked at the ends of these segments. The new endpoints have been connected to form the third side of a larger triangle, and equally spaced points have been constructed along that side. `routine triangle4 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Continue the process for one more step. How many points do you have in the new triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 21 points confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You will add an endpoint to each newly extended side, so each of the new sides will contain 6 points. You will then have to add 4 equally spaced points to the new side, giving you a total of 15 points on the new triangle. There are also 5 marked points inside the triangle for a total of 21 marked points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. The sequence of marked points is 3, 6, 10, 15, 21. What do expect will be the next number in this sequence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 21+7= 28 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The differences between these numbers are 3, 4, 5, 6. The next difference, according to this pattern, should be 7, which would make the next number 28. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. How can you tell, in terms of the process you used to construct these triangles, that the next number should be 7 greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To make the triangle bigger again you add2 more points making 7 mew points confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When you extend the triangle again, you will add two new endpoints and each side will now have 7 points. The 7 points on the new triangle will be all of the new points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. How do you know this sequence will continue in this manner? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You just add one point to the ends each time confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Each time you extend the triangle, each side increases by 1. All the new marked points are on the new side, so the total number of marked points will increase by 1 more than with the previous extension. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Math 151 007. `Query 7
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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: ********************************************* Question: `q1.2.18 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4^2+4=5^2-5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The next equation in the sequence would be 4^2 + 4 = 5^2 - 5 The verification is as follows: 4^2 + 4 = 5^2 - 5 simplifies to give you 16 + 4 = 25 - 5 or 20 = 20 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The square of the sum of the first number is equal to the sum of the numbers cubed confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q1.2.36 1 st triangular # div by 3, remainder; then 2d etc. Pattern. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** " 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!