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• 1. Investigate:1 1 1 3 5 9 17 31 57 105 a. Find the next three terms: b. Write a mathematical sentence which can be used as the formula in finding the nth term of the sequence. 1 1 1 3 5 9 17 31 57 105 193 355 653 a. The next three terms are as follows, 193 for the 11th term, 355 for the 12th term, and 653 for the 13th term. b. If we are going to investigate this 3 9 31 105 193 355 653 11th 12th 13th 5 17 57
• 2. sequence, we can observe that for each number in the series (starting from the value 3 [4th term] to last) is the sum of preceding 3 numbers. We can deduce that this is a Tribonacci Sequence. The Tribonacci sequence starts with three predetermined terms and then each successive term is the sum of the three preceding terms. The table below shows the observation: What to find (value) How to find 4th term (3) 1st + 2nd + 3rd terms (1+1+1) 5th term (5) 2nd + 3rd + 4th terms (1+1+3) 6th term (9) 3rd + 4th + 5th terms (1+3+5) 7th term (17) 4th + 5th + 6th terms (3+5+9)
• 3. In order to solve for the next three terms (11th, 12th and 13th), we will make use of this recurrence equation on finding for the nth term where n ≥ 4: Example, this is the formula if we want to find the 4th term of a Tribonacci sequence. T4=T1+T2+T3 which comes from the equation Tn= Tn−3+ Tn−2+Tn−1 (the formula for 8th term (31) 5th + 6th + 7th terms (5+9+17) 9th term(57) 6th + 7th +8th terms (9+17+31) 10th term(105) 7th +8th +9th terms (17+31+57) 11th term (193) 8th +9th + 10th terms (31+57+105) 12th term (355) 9th + 10th + 11th terms (57+105+193) 13th term (653) 10th +11th +12th terms (105+193+355)
• 4. finding the nth term of a tribonacci sequence)
• 5. 3. For this kind of tile sequence, we can assert that we can have different approach for solving the nth figure. In our way, we used a function and figures to solve this kind of problem. First, we need to fill out each figures as a dimension of square. To do this, let us name first the shown sequence of tiles as figures Visualizing the above sentence, here are the figures formed: Figure 1 Figure 2 Figure 3 2x2 dimension 3x3 dimension 4x4 dimension
• 6. Figure 4 Knowing that every nxn dimension are added by 1x1 dimension to arrive for the next figure, we can conclude that the next figure would be a 6x6 dimension square. We can also explore from the figures that in order to arrive at the dimension, we can show it on mathematical expression (n+2)(n+2). Say, for example, for figure 1, we noticed that we 5x5 dimension
• 7. don’t need to substitute any natural numbers to get 2x2 dimension for the first figure. Therefore, with a view to get 2x2 dimension, we will just substitute zero on the expression. (0+2)(0+2) will be simplified to (2)(2) or 2x2 dimension. Now for the figure 2, we noticed that the tiles are in 3x3 form, therefore, in order to arrive for 3x3 dimension, we are going to add 1 shift horizontally and 1 shift vertically which then can be substituted to the expression which can be seen as (1+2)(1+2) => (3)(3) or 3x3 dimension. The pattern which can be noticed could be traced for the next figures. Now, we can see that the tiles without shades can be expressed into mathematical expression 2(n+1), (n+1) tiles for the first column and (n+1) tiles for the last column. To better understand, we can visually show the expression using the figures above. Say for example, for figure
• 8. 3: We arrived at the expression 2(n+1) to get the total number of tiles which are not shaded by blue. If we are going to construct a table for better understanding, follow below. n+1 2+1 3 non-shaded tiles for first column n+1 2+1 3 non-shaded tiles for last column 2(2+1) = 4+2 = 6 non- shaded tiles
• 9. Now, going back to the problem letter a. How many square tiles will it take to build the next figure? Since we know that the predicted dimension of the figure (which we’ll be calling figure 5) is 6x6. We can tell that there are 4 shifts added vertically and Figure Dimension Shift added vertically and horizontally (n) Mathematical expression (n+2)(n+2) Mathematical expression 2(n+1) 1 2x2 0 shift / No shift (0+2)(0+2) 2(0+1) 2 3x3 1 shift (1+2)(1+2) 2(1+1) 3 4x4 2 shifts (2+2)(2+2) 2(2+1) 4 5x5 3 shifts (3+2)(3+2) 2(3+1) 5 6x6 4 shifts (4+2)(4+2) 2(4+1)Next figure
• 10. horizontally inferring to the figure 1 which can be shown below. Now in, order to find the the non-shaded tiles, we will just substitute n=4 to 2(n+1) => 2(4+1) = 10 non-shaded tiles. Added shifts four units horizontally Added shifts four units vertically 6x6 dimension square
• 11. This basically means that there are 5 non- shaded tiles for top left column and 5 non- shaded tiles for bottom right column of a 6x6 dimension which can be shown below. We are asked to find the number of square tiles (shaded tiles). We can find the shaded tiles by inspection which is 26 shaded tiles. On the other side, we can also figure out the number of the shaded tiles by making a function shown below. n+1 4+1 5 non-shaded tiles for last column n+1 4+1 5 non-shaded tiles for first column
• 12. f(n) = (n+2)(n+2)-2(n+1) where (n+2)(n+2) will be the arrangement/dimension of the tiles and 2(n+1) are tiles which are non-shaded. Knowing that n=4, we can substitute it to the function; f(4) = (4+2)(4+2) – 2(4-1) = (6)(6) – 2(5) = 36 – 10 = 26 Therefore, we conclude that we can also get the number of shaded tiles by using functions. b. To get the number of tiles for an nth figure, we will subtract 2(n+1) from (n+2)(n+2) which can be shown as a function: Solving by Function = Solving by Inspection 26 = 26 TRUE
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