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1 WBHS Mathematics Grade 10 Algebra Patterns Investigation Memo– 17 Jan 2017 Marks: 45 Time: 1 hour 1. A calculator may be used in this investigation. 2. Part 1 and Part 2 must be detached and handed in with your script. NAME: ______________________________ MATHS TEACHER: ___________________ PART 1 (to be completed on this page, detached and handed in with your script) Below is a “number cell”. In a number cell, the first two numbers are added to get the third number. The second and third numbers are added to get the fourth etc 3 1.1. 4 7 11 18 12√ 19√ 31√ Complete the following number cell: 5 7 (3) 1.2. Complete the following number cell in which you are given only the first and last numbers. 12√ 8 20√ 32√ 52 (3) 1.3. Complete the following number cell: x x+ y 2x+y√ 3x+2y√ 5x+3y√ 8x+5y√ 13x+8y√ 21x+13y√ (6) 1.4. Describe fully the pattern that you observe in the coefficients of x and y. The x-co-efiicients: the first two numbers are added to get the third number. The second and third numbers are added to get the fourth etc. or it is the Fibonacci sequence.√ The y-co-efficients: follow the same pattern as for the x-co-efficients (Fibonacci).√ (2) [14] 2 PART 2 (to be completed on this page, detached and handed in with your script) The numbers 1 to 64 are arranged in an 8 × 8 grid below. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 A 3 × 3 square is drawn on the grid. The opposite corners are multiplied: 34 × 20 = 680 36 × 18 = 648 The difference is found: 680 – 648 = 32. 2.1 Using the same grid above, do this with two more examples. 22 x 8 = 176√ 54 x 40 = 2160√ 6 x 24 = 144√ 38 x 56 = 2128√ 176 – 144 = 32√ 2128 – 2160 = 32√ (Any 2 correct examples). 2.2 (any 2 correct egs.) (6) Each 8 x 8 has a general form of: x x+1 x+2 x+8 x+9 x+10 x+16 x+17 x+18 √ Now, (x + 16)(x + 2) = x² + 18x + 32√ And x(x + 18) = x² + 18x√ Therefore x² + 18x + 32 – (x² + 18x) = 32√ (3) [9] 3 PART 3 3.1 50cm 50cm 50cm50cm 100cm 50cm 50cm 100cm 100cm 50cm 50cm 50cm 100cm √ 50cm 100cm 100cm 100cm √ √ 100cm √ (4) 3.2 You add the 2 preceding number of pavers √to get the next amount of pavers√ (Fibonacci) (2) 3.3 89√√ (2) 4.1 Any 2 rows which are correct. (2) 4.2 A number squared plus one more than this number squared√ plus the product of these two numbers squared√ is equal to one more than the product of these two numbers squared.√ (3) 4.3 n2 + (n + 1)2 Ö + (n(n + 1))2 Ö = (n(n + 1) + 1)2 Ö 4.4 RTP: (3) 4 n 2 + (n + 1) 2 + (n(n + 1)) 2 = (n(n + 1) + 1) 2 Ö LHS = n 2 + n 2 + 2n + 1 + n 4 + 2n3 + n 2 = n 4 + 2n3 + 3n 2 + 2n + 1 Ö RHS = (n 2 + n + 1) 2 Ö (4) = n 4 + n3 + n 2 + n3 + n 2 + n + n 2 + n + 1 = n 4 + 2n3 + 3n 2 + 2n + 1 Ö \ LHS = RHS 4.5 192 + 202 + 3802 = 3812 Ö Ö (2)