Chapter : Patterns and Relationships Getting Started, p. 13 1. a) The factors of 1 are 1,, 3,, 6, and 1. The factors of are 1,,, 7, 1, and. The greatest common factor is. b) The factors of 16 are 1,,,, and 16. The factors of 6 are 1,, 3,, 5, 6,, 1, 15, 3, and 6. The greatest common factor is. c) The factors of are 1,,, 11,, and. The factors of are 1,, 11, and. The factors of are 1,,,, 11,,, and. The greatest common factor is. d) The factors of 95 are 1, 5, 19, and 95. The factors of 5 are 1, 5, and 5. The factors of 55 are 1, 5, and 11. The greatest common factor is 5.. a) Figure number Figure Number of counters 1 3 5 3 7 9 5 11 b) Add counters to the middle row of the previous figure to make the new figure. c) If n represents the figure number, then the number of counters is given by n + 1. d) For the th figure, use n =. So, () + 1 = 1 counters would be needed. 3. a) 3, 3 + 3 = 6, 6 + 3 = 9, 9 + 3 = 1, 1 + 3 = 15, 15 + 3 = 1, 1 + 3 = 1. The pattern rule is start at 3 and add 3 each time. b), + = 6, 6 + =, + = 1, 1 + = 1, 1 + =, + = 6. The pattern rule is start at and add each time. c) 1, 1 + = 3, 3 + 3 = 6, 6 + =, + 5 = 15, 15 + 6 = 1, 1 + 7 =. The pattern rule is start at 1 and add, then add 1 more each time. d) 1,, 9, 16, 5, 36, 9. There are two possible pattern rules. i) Start at 1 and add the next odd number each time. 1, 1 + 3 =, + 5 = 9, 9 + 7 = 16, 16 + 9 = 5, 5 + 11 = 36, 36 + 13 = 9, ii) The sequence is the squares of numbers, starting at 1 and increasing by 1 each time. 1 = 1, =, 3 = 9, = 16, 5 = 5, 6 = 36, 7 = 9,. a) n + b) 5n + c) 3n d) n 1 5. a) 15 a = 15. =. b) 3b + = (3 1) + = + = c) 6b 5 = (6 1) 5 = 5 = 79 d). + a =. + (.) =. + 19. = 6. a) The number of blue tiles stays the same and the number of orange tiles increases by each time. b) The pattern rule is double the figure number and add 1. The add 1 is for the blue block that is always the same. c) n + 1, where n represents the figure number. d) For figure 5, use n = 5. n + 1 = 5 + 1 = + 1 = 11 There will be 11 square tiles in figure 5. 7. a) Plot the points (1, 15), (, 3), (3, 5), and (, 6). 15 135 5 9 75 6 5 3 15 1 Savings Compared to Number of Weeks 3 5 6 Number of weeks 7 9 Nelson Mathematics Solutions -1
b) Connect the points. They form a straight line. The line passes through the point for Week 9 at $135. She will have saved $135. c) The line passes through the point for $9 at Week 6. She will have saved $9 in Week 6.. Creating Pattern Rules from Models, pp. 1 19 3. a) The vertical column of three blue tiles stays the same in each figure. There are two red tiles added to each new figure. b) An expression for the number of tiles is n + 3, where n is the figure number.. For example, the two vertical toothpicks are the same in all the figures. In each figure, a triangle made from 3 toothpicks is added on top. The pattern rule is 3n +. Another way to describe the pattern is that there are two vertical toothpicks that stay the same. In each figure, a left diagonal, a right diagonal, and a base are added. The pattern rule is + n + n + n. 5. a) In the first model, the three horizontal red tiles stay the same. Two purple tiles are added each time under the first and third red tiles. In the second model, the yellow tile stays the same. Two blue tiles are added each time. These new tiles are added in the column to the left and right of the yellow tile. b) For the first model: n + 3 For the second model: (n + 1) + 1 6. The pattern rule 3n + 1 describes the toothpick pattern. Each figure has 3 more toothpicks added. n and n 1 describe patterns where toothpicks are added each time. n + describes a pattern where toothpicks are added each time. Only 3n + 1 (choice C) describes the correct pattern. 7. The pattern is 3, 5, 7, 9,. Since the pattern increases by each time, the pattern rule has to include n. But each term value is too low by 1. So add 1. The pattern rule is n + 1. Another pattern rule is n + n + 1.. For example, n + and + n +. 9. a) In each figure, the number of red tiles is twice the figure number, so a pattern rule is n. b) For example, the left and right columns do not change. Each column is four tiles long for a total of tiles. In the middle columns the first figure had blue tiles, the second had blue tiles, and the third had 6 blue tiles. This can be expressed algebraically as n. Combine the two results to get the expression n +.. a) No. The number of blue tiles is expressed by n +. This expression is always even, since you keep adding multiples of to. b) The number of red tiles is expressed by n, where n represents the figure number, so solve n = to find the figure number. If n =, then n = 5. So, figure 5 contains red tiles. The number of blue tiles is expressed by + n. When n = 5, the number of blue tiles is: + (5) = + = 1 The figure with red tiles has 1 blue tiles. 11. For example, in the figures, the top three tiles stay the same. In figure 1, there is one tile beneath the three top tiles; in figure, there are four tiles beneath the three top tiles; and in figure 3, there are 9 tiles beneath the three top tiles. The pattern of the tiles beneath the three top tiles can be expressed by squaring the figure number. The three top tiles are always the same and can be expressed by 3. The pattern of the tiles beneath the three top tiles can be expressed as n. Combining the two expressions gives n + 3. 1. a) n + Figure 1 Figure Figure 3 b) n + Figure i 1 Figure Figure 3 c) n + n Figure 1 Figure Figure 3 13. a) In each figure, the number of red tiles is the square of the figure number, so a pattern rule is n. b) There are always blue corner tiles. Look at the pattern between the corner tiles. Figure 1 has tiles, figure has tiles, and figure 3 has 1 tiles. The number of tiles is always times more than the figure number. This can be expressed algebraically as n. Add the blue corner tiles to get n +. - Chapter : Patterns and Relationships
c) The total number of tiles is equal to the total number of red tiles plus the total number of blue tiles. Combine the answers from parts a) and b). n + (n + ) = n + n +. Notice that the pattern of the total number of tiles is the square of the whole numbers, starting at 3. This can be expressed algebraically as (n + )..3 The General Term of a Sequence, pp. 13 133 3. a) Term number (figure number) Picture Term value (number of squares) 1 3 6 5 b) In each figure, there are twice as many squares as the figure number, so an expression is n. c) Use n = 3. n = (3) = 6 The 3th term is 6.. a) Term number Term value 1 6 11 3 16 1 5 6 b) The term values increase by 5 each time. Repeated addition is the same as multiplication, so the pattern rule has to involve 5n. By itself, 5n generates the terms 5,, 15,, 5, and so on. These values are 1 less than the sequence in the question, so add 1 to get the expression 5n + 1. c) To calculate the value of the 5th term in the sequence, evaluate 5n + 1 for n = 5. 5n + 1 = 5(5) + 1 = 15 + 1 = 16 The 5th number in the sequence is 16. 5. Hendryk is right. The term values increase by each time. Repeated addition is the same multiplication, so the pattern rule has to involve n. By itself, n generates the terms,, 6,, and so on. These values are 3 less than the term values in the table, so add 3 to n. The correct pattern rule is n + 3. Nilay s pattern rule is 3n +, which describes a pattern that increases by 3 each time. 6. a) The term values are, 11, and 1. Each term increases by 3. Repeated addition is the same as multiplication, so the pattern rule has to involve 3n. By itself, 3n generates the terms 3, 6, 9, 1, and so on. These values are 5 less than the values in the pattern, so add 5 to 3n. The pattern rule is 3n + 5. b) For n =, 3n + 5 = 3() + 5 = + 5 = 5 For the th figure, 5 counters are needed. c) Try guess and test. Guess values for n that make 3n + 5 equal to 1. Try n = because 3n is equal to 3. So 3n + 5 = 3 + 5 = 35 This answer is 6 less than 1, so try n = 1. 3n + 5 = 3(1) + 5 = 36 + 5 = 1 This answer is correct. Figure 1 has 1 counters. 7. Figure 1 contains triangles, figure contains 9 triangles, and figure 3 contains 16 triangles. Notice that these values are squares of whole numbers: =, 3 = 9, and = 16. However, the pattern rule is not n, because the first figure is, not 1. Adjust the pattern rule by adding 1 to n to get (n + 1). Check that this pattern rule is right. For n = 1, (n + 1) = (1 + 1) = () = This answer is correct. For n =, (n + 1) = ( + 1) = (3) = 9 This answer is correct. For n = 3, (n + 1) = (3 + 1) = () = 16 This answer is correct. Nelson Mathematics Solutions -3
The pattern rule is (n + 1). Check Vanya s claim that the 15th figure (when n = 15) contains 5 triangles. For n = 15, (n + 1) = (15 + 1) = (16) = 56 This means that Vanya is incorrect, as 56 small triangles are in the 15th figure. Vanya most likely used n as the general term, not (n + 1). It is the 1th figure.. a) Start with counters and add more counters each time. b) counters are added each time. Repeated addition is the same as multiplication so the pattern rule involves n. By itself, n generates the terms,, 1, 16, and so on. These values are always 6 less than the actual number of counters, so add 6 to n to get the pattern rule n + 6. c) Use guess and test. Guess values of n that make the pattern rule n + 6 equal to 5. For n =, n + 6 = () + 6 = 6 This is too low by. For n = 11, n + 6 = (11) + 6 = 5 This answer is correct. Figure 11 has 5 counters. d) For n = 75, n + 6 = (75) + 6 = 3 + 6 = 36 Figure 75 has 36 counters. 9. a) The term values increase by each time. Repeated addition is the same multiplication, so the pattern rule involves n. By itself, n generates the terms,, 1, 16, and so on. These values are 1 less than the values in the sequence so add 1 to n. The expression is n + 1. For n = 5, n + 1 = (5) + 1 = + 1 = 1 The 5th value in the sequence is 1. b) n + 9. For n = 5, n + 9 = (5) + 9 = + 9 = 9 c) 5n + 1. For n = 5, 5n + 1 = 5(5) + 1 = 5 + 1 = 71 d) 3n. For n = 5, 3n = 3(5) = 15 = e) n + 1. For n = 5, n + 1 = (5) + 1 = 5 + 1 = 51. a) The numbers in the 5th row are 5,, 15,, 5,. When 3 is added to each number, the results are, 13, 1, 3,, b) Since we are starting in the fifth row, each term is a multiple of 5, so 5n is involved. This gives the terms 5,, 15, and so on, which is 3 less than our terms since we added 3 to each number. So, the algebraic pattern rule is 5n + 3. c) For example, add 3 to the 7th row of the multiplication table:, 17,, 31,. The pattern rule is 7n + 3. d) For example, all the sequences are multiples of a certain number plus a constant. For question b), the algebraic pattern rule 5n + 3 represents the multiples of 5, with the constant 3 added. For question 9c), the algebraic pattern rule 5n + 1 represents the multiples of 5, but with the constant 1 added. The pattern rule 3n from question 9d) can be thought of as the 3th row of an extended multiplication table minus. 11. a) The sequence is: 7,, 15, Since the terms increase by the same amount, is halfway between 7 and 15, so must be 11. The pattern rule is start at 7 and add each time. The sequence is 7, 11, 15, 19, 3, 7, 31. The 7th term is 31. An algebraic expression for the general term is n + 3. b) The sequence is,, 7,,,, 15, The terms increase by the same number each time, so they must increase by each time, starting at 3: 3, 5, 7, 9, 11, 13, 15, 17, 19, 1, 3, 5, 7, 9, 31, 33, 35. The 17th term is 35. An expression for the general term is n + 1. 1. a) Since the terms of the sequence decrease by each time, the pattern rule will include the expression n. n is off by 31 for each term, so add 31 to get the pattern rule n + 31. For the 5th term, or n = 5, n + 31 = (5) + 31 = + 31 = 69 The 5th term is 69. - Chapter : Patterns and Relationships
b) Since the terms of the sequence decrease by 1 each time, the pattern rule will include n. n is off by 119 for each term, so add 119 to get the pattern rule n + 119. For the 5th term, or n = 5, n + 119 = 5 + 119 = 69 The 5th term is 69. 13. a) Make a table of values and extend the table. Term number Term value Change in term value 1 7 11 + 3 16 +5 +6 5 9 +7 A rule for this pattern is start at 7 and add. Then add 1 more each time. If you were to re-arrange the dots you could form triangles with one dot below them. An algebraic expression for the nth triangular number n(n +1) is. However, this expression only works for triangular numbers if you start at 1. Since the first figure has a triangle with three rows, this is the third triangle number. You can fix the formula by adding (n + )(n + 3) to n, which gives. Add the left over (n + )(n + 3) counter to the pattern rule, to get +1. b) For n =, (n + )(n + 3) ( + )( + 3) +1 = +1 = ()(3) +1 56 = +1 = 553+1 = 55 c) For example, like in a), rearrange the figure so that a triangle is present in each figure, with a leftover counter at the bottom. Notice that each row in the triangle has one more counter than the previous row. The first row has 1 counter, the second has, and so on. The first figure has 3 rows and a leftover counter, and one more row is added each time, so the th figure will have rows and one leftover counter. Now add up the counters in all rows. This is the same as adding all the numbers from 1 to. Determine the sum of any pair by adding the first and last numbers, 1 + = 3. Since there are 5 pairs from 1 to, there will be 51 pairs from 1 to. Since each pair sums to 3, the sum of the numbers from 1 to is equal to 51 3 = 553. This means that for the th figure, there are 553 counters in the triangle. Add the leftover counter to get the total number of counters: 553 + 1 = 55. 1. a) Figure number Number of cubes 1 1 3 3 6 5 15 b) This pattern is the same as triangles with another layer added, except the triangles are rotated. So the n(n +1) pattern is still. c) For n =, n(n +1) = ( +1) = (11) = 1 = 55 15. a) 1,, 3,, 5, 6,... b) Yes. Detach all the rows from each other. Then pair up the top and bottom rows (these can be attached to form a new row). Next pair up the next-to-top and next-to-bottom rows. Continue until all the rows are paired up or until there is one row left over. Ignore the middle row for now. Multiply the number of rows by the number of cubes in one row. If another half of a row is needed, it can be added to find out the total number of cubes. Nelson Mathematics Solutions -5
Another way to phrase this is that the total number of cubes is equal to the number of cubes in one row multiplied by the number of rows and a half row. Mid-Chapter Review, p. 135 1. a) The sequence starts with two s, then every number is the sum of the previous two numbers. It is like the Fibonacci sequence, but each term is twice the matching term in the Fibonacci sequence. b) No, 97 could not be in this sequence because it is odd. Since an even number plus an even number is always even, every number in this sequence will be even. c) =, =, = = 16, 6 = 1, 16 1 = 6 = 36, =, 36 = =, 6 16 = 96, 96 = 16 = 56, 6 = 6, 56 6 = The difference alternates between and for each group of three numbers. So it is valid for other terms in the sequence. d) In the Fibonacci sequence, the difference alternates between 1 and 1. In this sequence, the difference alternates between and.. a) Each figure has a constant column of 5 tiles. Then, the first figure has one extra column of 3 tiles, the second figure has two extra columns of 3 tiles, and so on. The nth term is 3n + 5. b) The 5 represents the column of five green tiles and the 3n represents the group of columns of three orange tiles. 3. a) You could colour the three circles on the left one colour, and the rest another colour: Or, colour each row one colour: b) In the first model, the nth term contains 3n + 3 circles. In the second model, the nth term contains n + (n + 1) + (n + ) circles.. a) Each number is 3 more than the number before it, so the algebraic expression must contain 3n. The values are less so add. An expression is 3n +. b) Each number is 3 more than the number before it, so the algebraic expression must contain 3n. The values are 5 less so add 5. An expression is 3n + 5. c) The second algebraic expression is equivalent to 1 more than the first algebraic expression. 5. a) Each number is more than the number before it, so the algebraic expression must contain n. The values are 3 less so add 3. An expression is n + 3. b) The 5th term is n + 3 = (5) + 3 = + 3 = 3 c) Guess the th term: () + 3 = 3. This is just less than 5, so try the 11th term: 3 + (11) = 5. The 11th term is 5. 6. a) Each number is 6 more than the number before it, so the algebraic expression must contain 6n. = 6 +, so try 6n +. It works for the other numbers also. The th term is 6() + = 6 + = 6 b) Each number is 3 more than the number before it, so the algebraic expression must contain 3n. 15 = 1 + 3, so try 3n + 1, and it works for the other numbers also. The th term is 3() + 1 = 3 + 1 = 31 c) Each number is 11 more than the number before it, so the algebraic expression must contain 11n. Without adding anything else, it works for the other numbers also. The th term is 11n = 11() = 1 d) Each number is 1 more than the number before it, so the algebraic expression must contain 1n, or n. 9 = 1 +, so try n +. It works for the other numbers also. The th term is n + = + = e) Each number is 16 more than the number before it, so the algebraic expression must contain 16n. 7 = 16 + 31, so try 16n + 31. It works for the other numbers also. The th term is 16n + 31 = 16() + 31 = 16 + 31 = 1631-6 Chapter : Patterns and Relationships
. Solve Problems by Examining Simpler Problems, p. 13. Try to find a pattern relating the number of cannonballs in one layer to the number of cannonballs in the next layer. The width and length of the next layer will always be one larger than the width and length of the previous layer. Try finding a pattern by examining this table. Layers Dimensions of layer Balls in bottom layer 1 1 3 3 3 3 5 15 6 n n (n + ) The number of balls is the product of the dimensions of the layer. The dimensions increase by one in the next layer. The 9th layer will have dimensions of 9 11, and will require 99 balls. 5. First look at the number of pieces after the first few cuts. After starting with one full piece, there are 3 pieces, 5 pieces, 7 pieces, and so on. Every cut creates two more pieces of spaghetti. The number of pieces after n cuts can be represented by n + 1. After cuts, there will be ( ) + 1 = 1 pieces. 6. a) Suppose there are a smaller number of teams and they each play all the other teams twice. How many games will be played? Organize this information into a table to see if any patterns become apparent. Number Pattern Games Change of teams played 1 1 1 + 3 3 6 + 3 1 +6 To find the next term, add the next even number. Another way is to think of each team playing a home game against every other team. This means that any given pair of teams will have played games against each other at the end of the season. If there are n teams in the league, each teams plays (n 1) other teams, so there are n(n 1) games played in total. b) You can extend the table for more values up to 1 teams, or you can use your algebraic expression and replace n with 1. When the league has 1 teams, 1 13 = 1 games will be played. 7. For a simpler problem, ignore hose C and find how long it takes hose A and hose B to fill the pool. If they run together for 6 h, then they will fill 3 pools. Hose A fills 1 pool and hose B fills pools. If the hoses can fill 3 pools in 6 h, then they can fill a single pool in h. Reintroduce hose C back into the problem and use the same method. If all 3 hoses run for 6 h, then hose A can fill one pool, hose B can fill pools, and hose C can fill 3 pools. Together they can fill 6 pools in 6 h, or a single pool every hour. It takes one hour to fill the pool.. a) Solve the simpler problem with 1 pennant, then pennants, then 3 pennants. If you have one pennant, then you have to put the pennant in the first spot. There is one possibility (1). If you have two pennants, then you have two choices for the first spot (either pennant 1 or pennant ). Once you make your choice, you will have one pennant left, and this pennant must go in the second spot. There are choices in all. If you have 3 pennants, then you have 3 choices for the first spot. Once you make that choice, you have two pennants left for spots. This now becomes the same as the problem with only two pennants that you just solved. For each choice in the first spot (3 choices), there are then pennants, which can be placed in different ways. This means there are 3 outcomes. Notice that as you add more pennants, you can still use the work you ve done before. Summarize your work in a table. Number of Choices Possible pennants arrangements 1 1 1 1 3 3 1 6 You see the pattern for the choices is to multiply together all the whole numbers less than or equal to the number of pennants. This means that the number of ways in which pennants can be arrange is 3 1 =. b) Follow the pattern from part a). The number of ways in which 5 pennants can be arranged is 5 3 1 =. c) Follow the pattern from part a). The number of ways in which pennants can be arranged is 9 7 6 5 3 1 = 3 6. Nelson Mathematics Solutions -7
9. The width of each rectangle is the same as the figure number. The length is one more than the figure number. The nth rectangle is n units by (n + 1) units. The area is the product of these two side lengths. The th rectangle is units by 1 units, so its area is Area = units 1 units = square units. Questions in parts a), b), and c) can be answered by creating a table of values and looking for patterns. a) b) c) Figure Panes with sides touching frame Panes with 1 side touching frame Panes with no sides touching frame 1 = 1 1 = 1 = = 3 1 = 3 9 = 3 n n n = = 11. Create a table of values and look for a pattern. Number of Number of Change sides diagonals 5 5 add 3 6 9 add 7 1 add 5 To find the number of diagonals, you need to add two less than the number of sides of the polygon to the number of diagonals of the previous polygon. Another way to look at this problem is figuring out how the diagonals are drawn. From every point, a line is drawn from itself to every other point on the polygon except itself and its directly adjacent points. The total number of diagonals that can be drawn in this way is n(n 3). One problem with this method is that it will create two copies of each diagonal, that is, one from A to G, and another from G to A. Since AG and GA are the same diagonals, divide the expression by, to get n(n 3). For example, consider a pentagon. B A E D C Draw the possible diagonals from the perspective of every vertex B B B A A E E B D D C C A A E E B For each of the five vertices, there are two possibilities. This corresponds to the n(n 3) part of the formula, because if you substitute five for n, you get 5(5 3) = 5() = However, some of the diagonals are the same. For example, AC in the first figure is the same as CA in the third figure. Note that every diagonal has an identical pair. But these pairs, for example AC and CA, are the same. Since there are twice as many diagonals, you have to divide the formula by to get n(n 3). If a polygon has sides, it will have a total of ( 17) = 17 diagonals. 1. For example, how many counters are used in the 11th figure of this pattern? Try finding a pattern in the number of counters in the first four diagrams. Figure number Number of counters 1 1 3 9 16 The number of counters used is n. Following this pattern, figure number 11 will use 11 = 11 counters in total..5 Relating Number Sequences to Graphs, pp. 1 13 3. The coordinates for Pattern A are (1, 6), (, 11), (3, 16), (, 1), and (5, 6). Pattern A matches the red scatter plot. D D C C A E D C - Chapter : Patterns and Relationships
Pattern B is n. The coordinates are (1, ), (, ), (3, 1), (, 16), (5, ), and (6, ). Pattern B matches the blue scatter plot. Pattern C is n + 3. The coordinates are (1, 5), (, 7), (3, 9), (, 11), (5, 13), and (6, 15). Pattern C matches the green scatter plot.. a) For example, create a table of values. Baskets picked Earnings ($) 5 5 Graph the points with the coordinates (, ), (, 5), (, ), and (, 5). Connect these points and extend the line. 36 3 16 Strawberry Earnings 6 Baskets picked 16 b) Draw a horizontal line from $36 on the Earnings axis to the extended line. It meets the line at 1 baskets. This means that he has to pick 1 baskets to earn $36. c) If b represents the number of baskets picked, an algebraic expression describing the relationship the earnings for number of baskets picked is.5b. 5. a) The coordinates are (1, 3), (, 6), and (3, 9). This is the blue graph. b) The coordinates are (1, 5), (, ), and (3, 11). This is the green graph. c) The coordinates are (1, 1), (, ), and (3, 7). This is the red graph. 6. a) Draw a scatter plot of volume related to time. The volume is L at min and the volume is L at min, so two points have the coordinates (, ) and (, ). Plot these points. Since the tank empties at a constant rate, you can connect them with a straight line. 15 5 Volume Compared to Time 3 5 6 Time (min) 7 9 b) Draw a horizontal line from 6 L on the Volume axis to the line connecting the two points. It meets the line at the 7 min mark. So, the volume is 6 L at 7 min. c) An expression for volume is t +, where t is time in minutes. 7. a) Graph the points with the coordinates (, 5), (1,.5), (, ), and (6, ). Connect these points with a straight line. 16 1 1 Evaporation 3 Time (h) b) Draw a horizontal line from 17.5 ml on the Volume axis to this line. It meets the line at the time of 3 h. So, the volume is 17.5 ml when 3 h have passed. c) The volume decreases by.5 ml every hour. An expression for the volume is.5n + 5, where n represents the number of hours.. a) 1 16 1 1 6 Term Value Compared to Term Number 5 6 7 1 3 Term number b) Let n represent the term number. The first pattern rule can be described as n. The second pattern rule can be described as n. 5 Nelson Mathematics Solutions -9
c) Both relationships have increasing values, and for the first few terms, the values are close. The second relationship is increasing by the same amount every term, that is in a straight line, while the first relationship is increasing faster and faster, in a curved line. 9. Term value 16 1 Term Value Compared to Term Number 1 3 5 6 7 Term number The pattern does not create a straight line on the graph. It is difficult to predict the 11th triangular number using this graph, because it starts curving, and you don t know how much it will curve by the 11th term.. A: Each term value is the square of the term number, so an algebraic expression is n. B: Each term value is 1 less than the square of the term number, so the expression is n 1. C: Each term value is 1 more than the square of the term number, so the expression is n + 1. 11. The first term value is 6, and the second term value is 6 + 5 = 11. The third term value is 11 + 7 = 1, and the fourth term value is 1 + 9 = 7. Each term is the previous term plus the next odd integer. So the pattern of increase for the term value is 5, 7, 9, 11, and so on. This pattern is described by the algebraic expression n + 3. If you subtract these values from the term numbers, you get the sequence 6 5 = 1, 11 7 =, 1 9 = 9, 7 11 = 16, and so on. This sequence is described by n. Since these two sequences are applied to the term numbers together, you must add them together. The algebraic expression for this table of values is n + n + 3. Chapter Self-Test, p. 15 1. a) b) The two shaded squares remain the same throughout the pattern. The white squares form an n by n square. c) There are n white squares, and a constant extra squares, for a total of n + squares.. a) Term number Term value Change 1 7 1 +5 3 17 +5 +5 5 7 +5 6 3 +5 The term value increases by 5 each time. A pattern rule for describing this sequence is 5n +. b) The 5th term is 5(5) + = 15 + = 17 3. a) Each term increases by 7. To calculate any term, multiply the term number by 7 and then add. An algebraic expression for this pattern rule is 7t + b) The nth term is 7n +. c) The 77th term can be calculated by replacing n with 77 in the algebraic expression. 7(77) + = 539 + = 59 The 77th term is 59.. Parts a), b) and c) can be answered by creating a table of values and extending it to levels. a) 56 balls are needed. b) He would need 56 = 16 more balls to increase the number of levels to. c) From the table, he would have 9 levels if he used 165 balls. Level Balls on level Change Total balls 1 1 1 3 + 3 6 +3 + 5 15 +5 35 6 1 +6 56 7 +7 36 + 9 5 +9 165 55 + - Chapter : Patterns and Relationships
5. For the first few cuts, the pattern is: Number of cuts Number of pieces 1 1 7 3 13 5 16 Each time a cut is made, the number of pieces increases by 3. Since it starts in one piece, an algebraic expression is 3n + 1. To find out how many pieces there are after 75 cuts, replace n with 75 in our algebraic expression. 3(75) + 1 = 5 + 1 = 6 There will be 6 pieces of spaghetti after 75 cuts. 6. a) For i), plot the points with coordinates (, 6), (5,, (7, 16), and (, ). For ii), plot the points with coordinates (6, 1), (, 3), (1, ), and (1, 7). Connect each set of points with a straight line and extend the line. 1 16 1 1 6 Term Value Compared to Term Number i) 6 1 1 16 1 Term number b) For i), draw a horizontal line from on the Term value axis to meet the line. It meets where the term number is 3. For ii), draw a horizontal line from on the Term value axis to meet the line. It meets where the term number is. 7. a) Since it increases by 5 each time, the pattern rule has to include 5n. But each term value is too low by. So add. An algebraic expression is 5n +. b) Create a table of values. Term number Number of toothpicks 1 9 1 3 19 ii) Graph the points with the coordinates (1, 9), (, 1), (3, 19), and (, ). Connect the points and extend the line. 9 7 6 5 3 Number of Toothpicks Compared to Term Number 6 1 1 16 1 Term number Draw a horizontal line from 9 on the number of toothpicks axis to meet the extended line. It meets the line at the 1th term number, so figure 1 will have 9 toothpicks. c) Substitute n = 1 in the algebraic expression. 5n + = 5(1) + = 9 + = 9 This confirms the value from part b). Chapter Review, p. 17 1. a) The number of pennies shown on the arms radiating out from the centre increase by one every figure. The centre penny stays the same for each figure. For the first figure, there are five pennies around the centre penny. For the second figure, there are ten pennies around the centre penny. For the third figure, there are 15 pennies around the centre penny. Five more pennies are added each time. b) The first few terms are 6, 11, 16,, so if you multiply the term number by five and then add one, you get term value. c) An algebraic pattern rule is 5n + 1, representing the number of pennies in figure n. d) To determine how many pennies are needed for the 1th poster in the sequence, we can use our algebraic expression and replace n with 1. 5n + 1 = 5(1) + 1 = 6 + 1 = 61 A total of 61 pennies are needed for the 1th poster. Nelson Mathematics Solutions -11
. a) Term Term Change number value 1 5 7 + 3 9 + 11 + The term values increase by each time. b) Since the pattern increases by each time, the pattern rule has to include n. But each term value is too low by 3. So add 3. The pattern rule is n + 3. c) The 7th term is n + 3 = (7) + 3 = + 3 = 13 3. The first figure has 6 counters and then 3 counters are added for every new figure. The algebraic expression for this sequence is 3n + 3. The 5th term will have 3(5) + 3 = 153 counters.. Use a table. Start with rays and increase. Number of Number of Change rays angles 1 3 3 + 6 +3 5 + 6 15 +5 7 1 +6 +7 9 36 + There are 36 angles created when 7 extra rays are drawn in a right angle. 5. a) Plot the points and connect them with straight lines. 1 16 1 1 6 1 Term Value Compared to Term Number 3 5 6 7 Term number b) From the graphs, the missing term values are: i) 7, 1.5 ii) 19, 1 iii) 19, 1 c) i) The pattern rule includes 1 n since the pattern increases by 1. But each term value is too low by 1. So add 1. The pattern rule is 1 n +1. ii) The pattern rule includes n since the pattern increases by. But each term value is too high by 1. So subtract 1. The pattern rule is n 1. iii) The pattern rule includes n since the pattern increases by. But each term value is too low by 3. So add 3. The pattern rule is n + 3. iii) 9 ii) i) 11 1-1 Chapter : Patterns and Relationships