{"id":171,"date":"2024-07-01T13:43:30","date_gmt":"2024-07-01T13:43:30","guid":{"rendered":"https:\/\/routledgelearning.com\/teachingsecondarymathematics\/?post_type=content&p=171"},"modified":"2024-08-07T10:36:14","modified_gmt":"2024-08-07T10:36:14","slug":"chapter-5-skills-in-teaching-mathematics","status":"publish","type":"content","link":"https:\/\/routledgelearning.com\/teachingsecondarymathematics\/content\/resources\/chapter-5-skills-in-teaching-mathematics\/","title":{"rendered":"Chapter 5 – Skills in Teaching Mathematics"},"content":{"rendered":"\n
How wonderful it would be if we knew everything about students, teaching, mathematics, and the teaching of mathematics. We could bottle it, sell it, become rich, and solve a lot of problems for everyone in the process. We all know there is no magic formula for teaching mathematics. In the process of attempting to find keys, we often stumble and miss objectives, but we also learn a variety of things that can be cataloged and used at a later date in other settings. This chapter deals with a broad collection of ideas, concepts, and strategies that should be useful in a variety of teaching environments. As you work through your career and attempt these suggestions, you will compile a list of methods and activities that will help your students learn mathematics.<\/p>\n<\/div>\n\n\n\n
Computational Madness<\/strong><\/p>\n\n\n\n\nSimplify the following expression.<\/li>\n<\/ol>\n\n\n\n\n \n 1234567890<\/mn>\n \n \n 1234567891<\/mn>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n (<\/mo>\n 1234567890<\/mn>\n )<\/mo>\n (<\/mo>\n 1234567892<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mfrac>\n<\/math>\n\n\n\nHint: Try using x = 1234567891<\/p>\n\n\n\nAnswer<\/summary>\nAnswer\/solution: 1234567890.<\/p>\n\n\n\nLet 1234567891 = N, then 1234567890 = N \u2013 1 and 1234567892 = N + 1.<\/p>\n\n\n\nTherefore,<\/p>\n\n\n\n\n \n \n \n \n \n 1234567890<\/mn>\n \n \n 1234567891<\/mn>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n (<\/mo>\n 1234567890<\/mn>\n )<\/mo>\n (<\/mo>\n 1234567892<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n <\/mrow>\n <\/mtd>\n <\/mtr>\n \n \n \n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n \n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n (<\/mo>\n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n )<\/mo>\n (<\/mo>\n N<\/mi>\n +<\/mo>\n 1<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n \n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n \n (<\/mo>\n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n 1<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n 1<\/mn>\n <\/mfrac>\n =<\/mo>\n 1234567890<\/mn>\n <\/mrow>\n <\/mtd>\n <\/mtr>\n <\/mtable>\n<\/math>\n<\/details>\n<\/div>\n\n\n\n\n\nBy arranging the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 it is possible to come up with a fraction equivalent to one-eighth. For example:<\/li>\n<\/ol>\n\n\n\n\n \n \n 1<\/mn>\n 8<\/mn>\n <\/mfrac>\n =<\/mo>\n \n 3187<\/mn>\n 25796<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\nYour task is to arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 to come up with an equivalent fraction to one-fifth<\/p>\n\n\n\nAnswer<\/summary>\nAnswer: There are many possibilities. One such answer is<\/p>\n\n\n\n\n \n \n 1<\/mn>\n 5<\/mn>\n <\/mfrac>\n =<\/mo>\n \n 52769<\/mn>\n 13845<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\nOthers include:<\/p>\n\n\n\n\n 2769 \/ 13845<\/li>\n\n\n\n2973 \/ 14865<\/li>\n\n\n\n9237 \/ 46185<\/li>\n\n\n\n2697 \/ 13485<\/li>\n\n\n\n2937 \/ 14685<\/li>\n\n\n\n2967 \/ 14835<\/li>\n\n\n\n3297 \/ 16485<\/li>\n\n\n\n3729 \/ 18645<\/li>\n\n\n\n6297 \/ 31485<\/li>\n\n\n\n9723 \/ 48615<\/li>\n\n\n\n9627 \/ 48135<\/li>\n\n\n\n7629 \/ 38145<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n\n\n\nAdditional Learning Activities<\/h2>\n\n\n\n\n\n1. <\/strong>Create a formula that shows how to determine the number of handshakes given to any number of people. Show that your formula works by adding a column to the table in the first exercise and completing it for each of the last three entries, showing your work.<\/p>\n\n\n\nYou might notice that these are the triangular numbers.<\/p>\n\n\n\nThe idea of generating formulas is common in the world of mathematics. We look at situations and try to come up with useful generalizations. Often these generalizations give us formulas that can be used in similar situations and that save time trying to find answers for each separate case. You have used many of these universally acceptable generalizations over your career. For example, formulas for area and perimeter, distances, rates, times, tax computations, and so on are among such generalizations.<\/p>\n\n\n\nYour Turn<\/strong><\/p>\n\n\n\n\nDevelop a formula for finding the sum of any set of consecutive counting numbers, beginning with any number. <\/li>\n<\/ul>\n\n\n\n\nGiven that n<\/mi> <\/mrow> (<\/mo> n<\/mi> <\/mrow> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac><\/math> gives the sum of the first n counting numbers, starting with 1, and that n2<\/sup> is the sum of the first n odd counting numbers, then it seems as if n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> \u2212<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup><\/math> should give the sum of the first n even counting numbers. <\/li>\n<\/ul>\n\n\n\n n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> −<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup> <\/mtd> <\/mi> =<\/mo> n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn>
Hint: Try using x = 1234567891<\/p>\n\n\n\nAnswer<\/summary>\nAnswer\/solution: 1234567890.<\/p>\n\n\n\nLet 1234567891 = N, then 1234567890 = N \u2013 1 and 1234567892 = N + 1.<\/p>\n\n\n\nTherefore,<\/p>\n\n\n\n\n \n \n \n \n \n 1234567890<\/mn>\n \n \n 1234567891<\/mn>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n (<\/mo>\n 1234567890<\/mn>\n )<\/mo>\n (<\/mo>\n 1234567892<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n <\/mrow>\n <\/mtd>\n <\/mtr>\n \n \n \n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n \n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n (<\/mo>\n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n )<\/mo>\n (<\/mo>\n N<\/mi>\n +<\/mo>\n 1<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n \n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n \n (<\/mo>\n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n 1<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n 1<\/mn>\n <\/mfrac>\n =<\/mo>\n 1234567890<\/mn>\n <\/mrow>\n <\/mtd>\n <\/mtr>\n <\/mtable>\n<\/math>\n<\/details>\n<\/div>\n\n\n\n\n\nBy arranging the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 it is possible to come up with a fraction equivalent to one-eighth. For example:<\/li>\n<\/ol>\n\n\n\n\n \n \n 1<\/mn>\n 8<\/mn>\n <\/mfrac>\n =<\/mo>\n \n 3187<\/mn>\n 25796<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\nYour task is to arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 to come up with an equivalent fraction to one-fifth<\/p>\n\n\n\nAnswer<\/summary>\nAnswer: There are many possibilities. One such answer is<\/p>\n\n\n\n\n \n \n 1<\/mn>\n 5<\/mn>\n <\/mfrac>\n =<\/mo>\n \n 52769<\/mn>\n 13845<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\nOthers include:<\/p>\n\n\n\n\n 2769 \/ 13845<\/li>\n\n\n\n2973 \/ 14865<\/li>\n\n\n\n9237 \/ 46185<\/li>\n\n\n\n2697 \/ 13485<\/li>\n\n\n\n2937 \/ 14685<\/li>\n\n\n\n2967 \/ 14835<\/li>\n\n\n\n3297 \/ 16485<\/li>\n\n\n\n3729 \/ 18645<\/li>\n\n\n\n6297 \/ 31485<\/li>\n\n\n\n9723 \/ 48615<\/li>\n\n\n\n9627 \/ 48135<\/li>\n\n\n\n7629 \/ 38145<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n\n\n\nAdditional Learning Activities<\/h2>\n\n\n\n\n\n1. <\/strong>Create a formula that shows how to determine the number of handshakes given to any number of people. Show that your formula works by adding a column to the table in the first exercise and completing it for each of the last three entries, showing your work.<\/p>\n\n\n\nYou might notice that these are the triangular numbers.<\/p>\n\n\n\nThe idea of generating formulas is common in the world of mathematics. We look at situations and try to come up with useful generalizations. Often these generalizations give us formulas that can be used in similar situations and that save time trying to find answers for each separate case. You have used many of these universally acceptable generalizations over your career. For example, formulas for area and perimeter, distances, rates, times, tax computations, and so on are among such generalizations.<\/p>\n\n\n\nYour Turn<\/strong><\/p>\n\n\n\n\nDevelop a formula for finding the sum of any set of consecutive counting numbers, beginning with any number. <\/li>\n<\/ul>\n\n\n\n\nGiven that n<\/mi> <\/mrow> (<\/mo> n<\/mi> <\/mrow> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac><\/math> gives the sum of the first n counting numbers, starting with 1, and that n2<\/sup> is the sum of the first n odd counting numbers, then it seems as if n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> \u2212<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup><\/math> should give the sum of the first n even counting numbers. <\/li>\n<\/ul>\n\n\n\n n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> −<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup> <\/mtd> <\/mi> =<\/mo> n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn>
Answer\/solution: 1234567890.<\/p>\n\n\n\n
Let 1234567891 = N, then 1234567890 = N \u2013 1 and 1234567892 = N + 1.<\/p>\n\n\n\n
Therefore,<\/p>\n\n\n\n\n \n \n \n \n \n 1234567890<\/mn>\n \n \n 1234567891<\/mn>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n (<\/mo>\n 1234567890<\/mn>\n )<\/mo>\n (<\/mo>\n 1234567892<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n <\/mrow>\n <\/mtd>\n <\/mtr>\n \n \n \n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n \n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n (<\/mo>\n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n )<\/mo>\n (<\/mo>\n N<\/mi>\n +<\/mo>\n 1<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n \n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n \n (<\/mo>\n \n N<\/mi>\n 2<\/mn>\n <\/msup>\n \u2212<\/mo>\n 1<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n \n \n N<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n <\/mrow>\n 1<\/mn>\n <\/mfrac>\n =<\/mo>\n 1234567890<\/mn>\n <\/mrow>\n <\/mtd>\n <\/mtr>\n <\/mtable>\n<\/math>\n<\/details>\n<\/div>\n\n\n\n\n\nBy arranging the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 it is possible to come up with a fraction equivalent to one-eighth. For example:<\/li>\n<\/ol>\n\n\n\n\n \n \n 1<\/mn>\n 8<\/mn>\n <\/mfrac>\n =<\/mo>\n \n 3187<\/mn>\n 25796<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\nYour task is to arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 to come up with an equivalent fraction to one-fifth<\/p>\n\n\n\nAnswer<\/summary>\nAnswer: There are many possibilities. One such answer is<\/p>\n\n\n\n\n \n \n 1<\/mn>\n 5<\/mn>\n <\/mfrac>\n =<\/mo>\n \n 52769<\/mn>\n 13845<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\nOthers include:<\/p>\n\n\n\n\n 2769 \/ 13845<\/li>\n\n\n\n2973 \/ 14865<\/li>\n\n\n\n9237 \/ 46185<\/li>\n\n\n\n2697 \/ 13485<\/li>\n\n\n\n2937 \/ 14685<\/li>\n\n\n\n2967 \/ 14835<\/li>\n\n\n\n3297 \/ 16485<\/li>\n\n\n\n3729 \/ 18645<\/li>\n\n\n\n6297 \/ 31485<\/li>\n\n\n\n9723 \/ 48615<\/li>\n\n\n\n9627 \/ 48135<\/li>\n\n\n\n7629 \/ 38145<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n\n\n\nAdditional Learning Activities<\/h2>\n\n\n\n\n\n1. <\/strong>Create a formula that shows how to determine the number of handshakes given to any number of people. Show that your formula works by adding a column to the table in the first exercise and completing it for each of the last three entries, showing your work.<\/p>\n\n\n\nYou might notice that these are the triangular numbers.<\/p>\n\n\n\nThe idea of generating formulas is common in the world of mathematics. We look at situations and try to come up with useful generalizations. Often these generalizations give us formulas that can be used in similar situations and that save time trying to find answers for each separate case. You have used many of these universally acceptable generalizations over your career. For example, formulas for area and perimeter, distances, rates, times, tax computations, and so on are among such generalizations.<\/p>\n\n\n\nYour Turn<\/strong><\/p>\n\n\n\n\nDevelop a formula for finding the sum of any set of consecutive counting numbers, beginning with any number. <\/li>\n<\/ul>\n\n\n\n\nGiven that n<\/mi> <\/mrow> (<\/mo> n<\/mi> <\/mrow> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac><\/math> gives the sum of the first n counting numbers, starting with 1, and that n2<\/sup> is the sum of the first n odd counting numbers, then it seems as if n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> \u2212<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup><\/math> should give the sum of the first n even counting numbers. <\/li>\n<\/ul>\n\n\n\n n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> −<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup> <\/mtd> <\/mi> =<\/mo> n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn>
Your task is to arrange the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 to come up with an equivalent fraction to one-fifth<\/p>\n\n\n\nAnswer<\/summary>\nAnswer: There are many possibilities. One such answer is<\/p>\n\n\n\n\n \n \n 1<\/mn>\n 5<\/mn>\n <\/mfrac>\n =<\/mo>\n \n 52769<\/mn>\n 13845<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\nOthers include:<\/p>\n\n\n\n\n 2769 \/ 13845<\/li>\n\n\n\n2973 \/ 14865<\/li>\n\n\n\n9237 \/ 46185<\/li>\n\n\n\n2697 \/ 13485<\/li>\n\n\n\n2937 \/ 14685<\/li>\n\n\n\n2967 \/ 14835<\/li>\n\n\n\n3297 \/ 16485<\/li>\n\n\n\n3729 \/ 18645<\/li>\n\n\n\n6297 \/ 31485<\/li>\n\n\n\n9723 \/ 48615<\/li>\n\n\n\n9627 \/ 48135<\/li>\n\n\n\n7629 \/ 38145<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n\n\n\nAdditional Learning Activities<\/h2>\n\n\n\n\n\n1. <\/strong>Create a formula that shows how to determine the number of handshakes given to any number of people. Show that your formula works by adding a column to the table in the first exercise and completing it for each of the last three entries, showing your work.<\/p>\n\n\n\nYou might notice that these are the triangular numbers.<\/p>\n\n\n\nThe idea of generating formulas is common in the world of mathematics. We look at situations and try to come up with useful generalizations. Often these generalizations give us formulas that can be used in similar situations and that save time trying to find answers for each separate case. You have used many of these universally acceptable generalizations over your career. For example, formulas for area and perimeter, distances, rates, times, tax computations, and so on are among such generalizations.<\/p>\n\n\n\nYour Turn<\/strong><\/p>\n\n\n\n\nDevelop a formula for finding the sum of any set of consecutive counting numbers, beginning with any number. <\/li>\n<\/ul>\n\n\n\n\nGiven that n<\/mi> <\/mrow> (<\/mo> n<\/mi> <\/mrow> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac><\/math> gives the sum of the first n counting numbers, starting with 1, and that n2<\/sup> is the sum of the first n odd counting numbers, then it seems as if n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> \u2212<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup><\/math> should give the sum of the first n even counting numbers. <\/li>\n<\/ul>\n\n\n\n n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> −<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup> <\/mtd> <\/mi> =<\/mo> n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn>
Answer: There are many possibilities. One such answer is<\/p>\n\n\n\n\n \n \n 1<\/mn>\n 5<\/mn>\n <\/mfrac>\n =<\/mo>\n \n 52769<\/mn>\n 13845<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\nOthers include:<\/p>\n\n\n\n\n 2769 \/ 13845<\/li>\n\n\n\n2973 \/ 14865<\/li>\n\n\n\n9237 \/ 46185<\/li>\n\n\n\n2697 \/ 13485<\/li>\n\n\n\n2937 \/ 14685<\/li>\n\n\n\n2967 \/ 14835<\/li>\n\n\n\n3297 \/ 16485<\/li>\n\n\n\n3729 \/ 18645<\/li>\n\n\n\n6297 \/ 31485<\/li>\n\n\n\n9723 \/ 48615<\/li>\n\n\n\n9627 \/ 48135<\/li>\n\n\n\n7629 \/ 38145<\/li>\n<\/ol>\n<\/details>\n<\/div>\n<\/div>\n\n\n\nAdditional Learning Activities<\/h2>\n\n\n\n\n\n1. <\/strong>Create a formula that shows how to determine the number of handshakes given to any number of people. Show that your formula works by adding a column to the table in the first exercise and completing it for each of the last three entries, showing your work.<\/p>\n\n\n\nYou might notice that these are the triangular numbers.<\/p>\n\n\n\nThe idea of generating formulas is common in the world of mathematics. We look at situations and try to come up with useful generalizations. Often these generalizations give us formulas that can be used in similar situations and that save time trying to find answers for each separate case. You have used many of these universally acceptable generalizations over your career. For example, formulas for area and perimeter, distances, rates, times, tax computations, and so on are among such generalizations.<\/p>\n\n\n\nYour Turn<\/strong><\/p>\n\n\n\n\nDevelop a formula for finding the sum of any set of consecutive counting numbers, beginning with any number. <\/li>\n<\/ul>\n\n\n\n\nGiven that n<\/mi> <\/mrow> (<\/mo> n<\/mi> <\/mrow> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac><\/math> gives the sum of the first n counting numbers, starting with 1, and that n2<\/sup> is the sum of the first n odd counting numbers, then it seems as if n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> \u2212<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup><\/math> should give the sum of the first n even counting numbers. <\/li>\n<\/ul>\n\n\n\n n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> −<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup> <\/mtd> <\/mi> =<\/mo> n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn>
Others include:<\/p>\n\n\n\n
1. <\/strong>Create a formula that shows how to determine the number of handshakes given to any number of people. Show that your formula works by adding a column to the table in the first exercise and completing it for each of the last three entries, showing your work.<\/p>\n\n\n\nYou might notice that these are the triangular numbers.<\/p>\n\n\n\nThe idea of generating formulas is common in the world of mathematics. We look at situations and try to come up with useful generalizations. Often these generalizations give us formulas that can be used in similar situations and that save time trying to find answers for each separate case. You have used many of these universally acceptable generalizations over your career. For example, formulas for area and perimeter, distances, rates, times, tax computations, and so on are among such generalizations.<\/p>\n\n\n\nYour Turn<\/strong><\/p>\n\n\n\n\nDevelop a formula for finding the sum of any set of consecutive counting numbers, beginning with any number. <\/li>\n<\/ul>\n\n\n\n\nGiven that n<\/mi> <\/mrow> (<\/mo> n<\/mi> <\/mrow> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac><\/math> gives the sum of the first n counting numbers, starting with 1, and that n2<\/sup> is the sum of the first n odd counting numbers, then it seems as if n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> \u2212<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup><\/math> should give the sum of the first n even counting numbers. <\/li>\n<\/ul>\n\n\n\n n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> −<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup> <\/mtd> <\/mi> =<\/mo> n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn>
You might notice that these are the triangular numbers.<\/p>\n\n\n\n
The idea of generating formulas is common in the world of mathematics. We look at situations and try to come up with useful generalizations. Often these generalizations give us formulas that can be used in similar situations and that save time trying to find answers for each separate case. You have used many of these universally acceptable generalizations over your career. For example, formulas for area and perimeter, distances, rates, times, tax computations, and so on are among such generalizations.<\/p>\n\n\n\n
Your Turn<\/strong><\/p>\n\n\n\n\nDevelop a formula for finding the sum of any set of consecutive counting numbers, beginning with any number. <\/li>\n<\/ul>\n\n\n\n\nGiven that n<\/mi> <\/mrow> (<\/mo> n<\/mi> <\/mrow> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac><\/math> gives the sum of the first n counting numbers, starting with 1, and that n2<\/sup> is the sum of the first n odd counting numbers, then it seems as if n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> \u2212<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup><\/math> should give the sum of the first n even counting numbers. <\/li>\n<\/ul>\n\n\n\n n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn> )<\/mo> <\/mrow> 2<\/mn> <\/mfrac> −<\/mo> n<\/mi> 2<\/mn> <\/mrow> <\/msup> <\/mtd> <\/mi> =<\/mo> n<\/mi> (<\/mo> n<\/mi> +<\/mo> 1<\/mn>