2D And 3D Sequences Project Essay Research
2D And 3D Sequences Project Essay, Research Paper
Plan of Investigation In this experiment I am traveling to necessitate the followers: A reckoner A pencil A pen Variety of beginnings of information Paper Ruler In this probe I have been asked to happen out how many squares would be needed to do up a certain form harmonizing to its sequence. The form is shown on the front page. In this probe I hope to happen a expression which could be used to happen out the figure of squares needed to construct the form at any sequencial place. Firstly I will interrupt the job down into simple stairss to get down with and travel into more item to explicate my solutions. I will exemplify to the full any methods I should utilize and explicate how I applied them to this certain job. I will foremost transport out this experiment on a 2D form and so widen my probe to 3D. The Number of Squares in Each Sequence I have achieved the following information by pulling out the form and widening upon it. Seq. no.12345678 No. Of cubes151325416185113 I am traveling to utilize this following method to see if I can work out some kind of form: SequenceCalculationsAnswer 1=11 22 ( 1 ) +35 32 ( 1+3 ) +513 42 ( 1+3+5 ) +725 52 ( 1+3+5+7 ) +941 62 ( 1+3+5+7+9 ) +1161 72 ( 1+3+5+7+9+11 ) +1385 82 ( 1+3+5+7+9+11+13 ) +15113 92 ( 1+3+5+7+9+11+13+15 ) +17 145 What I am making above is shown with the assistance of a diagram below ; If we take sequence 3: 2 ( 1+3 ) +5=13 2 ( 1 squares ) 2 ( 3 squares ) 1 ( 5 squares ) The Patterns I Have Noticied in Transporting Out the Previous Method I have now carried out ny first probe into the form and have seen a figure of different forms. First I can see that the figure of squares in each form is an uneven figure. Second I can see that the figure of squares in the form can be found out by taking the uneven Numberss from 1 onwards and adding them up ( harmonizing to the sequence ) . We so take the summing up ( & # 229 ; ) of these uneven Numberss and multiply them by two. After making this we add on the following back-to-back uneven figure to the twofold sum. I have besides noticied something through the drawings I have made of the forms. If we look at the symetrical sides of the form and add up the figure of squares we achieve a square figure. Trying to Obtain a Formula Through the Use of the Difference Method I will now use Jean Holderness & # 8217 ; difference method to seek and happen a expression. Pos.in seq. 123456 No.of squar. ( degree Celsius ) 1513254161 1st differ. ( a+b ) 048121620 2nd differ. ( 2a ) 44444 We can now utilize the equation an & lt ; sup & gt ; 2 & lt ; /sup & gt ; + bn + c & # 8216 ; n & # 8217 ; bespeaking the place in the sequence. If a = 2 so c = 1 and a + B = 0 If 2 is equal to b- so b = -2 I will now work out the equation utilizing the information I have obtained through utilizing the difference method: 1 ) 2 ( n -1 ) ( n & # 8211 ; 1 ) + 2n & # 8211 ; 1 2 ) 2 ( n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 2n + 1 ) + 2n & # 8211 ; 1 3 ) 2n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 4n + 2 + 2n & # 8211 ; 1 4 ) 2n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 2n + 1 Therefore my concluding equation is: 2n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 2n + 1 Proving My Equation and Using it to Find the Number of Squares in Higher Sequences I will now turn out my equation by using it to a figure of sequences and higher sequences I have non yet explored. Sequence 3: 1. 2 ( 3 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 6 + 1 2. 2 ( 9 ) & # 8211 ; 6 + 1 3. 18 -5 4. = 13 The expression when applied to sequence 3 appears to be successful. Sequence 5: 1. 2 ( 5 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 10 + 1 2. 2 ( 25 ) & # 8211 ; 10 + 1 3. 50 & # 8211 ; 10 + 1 4. 50 & # 8211 ; 9 5. = 41 Successful Sequence 6: 1. 2 ( 6 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 12 + 1 2. 2 ( 36 ) & # 8211 ; 12 +1 3. 72 & # 8211 ; 12 + 1 4. 72 & # 8211 ; 11 5. = 61 Successful Sequence 8: 1. 2 ( 8 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 16 + 1 2. 2 ( 64 ) & # 8211 ; 16 + 1 3. 128 & # 8211 ; 16 + 1 4. 128 & # 8211 ; 15 5. = 113 Successful The expression I found seems to be successful as I have shown on the old page. I will now utilize the expression to happen the figure of squares in a higher sequence. So now I wil use the expression 2n & lt ; sup & gt ; 2 & lt ; /sup & gt ; & # 8211 ; 2n + 1 to seek and happen the figure of squares contained in sequence 20. Sequence 20: 2 ( 20 & lt ; sup & gt ; 2 & lt ; /sup & gt ; ) & # 8211 ; 40 + 1 2 ( 400 ) & # 8211 ; 40 + 1 800 & # 8211 ; 40 + 1 800 & # 8211 ; 49 = 761 Alternatively of exemplifying the form I am traveling to utilize the method I used at the start of this piece of coursework. The method in which Iused to look for any forms in the sequences. I will utilize this to turn out the figure of squares given by the equation is right. As shown below: 2 ( 1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37 ) + 39 = 761 I feel this proves the equation to the full. Using the Difference Method to Find an Equation to Establish the Number of Squares in a 3D Version of the Pattern Pos.in seq.012345 No.of squar.-1172563129 1st differ.26183866 2nd differ.412202836 3rd differ.8888 So hence
we get the equation ; an? + bn2+ cn + vitamin D We already know the values of ‘n’ ( place in sequence ) in the equation so now we have to happen out the values of a, B, degree Celsius, and d. If n = 0 so d = -1 and if n = 1 so d = 1 I can now acquire rid of vitamin D from the equation to do it easier to happen the remainder of the values. I will will take n = 2 to make this in the undermentioned manner: 1st computation _ 8a + 4b + 2c + d a + B + c +d 7a + 3b + degree Celsius D will ever be added to each side of the equation. 2nd 8a + 4b + 2c = 8 = 4a + 2b + degree Celsiuss = 4 2 So so n = 2 8a + 4b + 2c = 8 = 4a + 2b + degree Celsiuss = 4 n = 3 27a + 9b + 3c = 26 n = 4 64a + 16b +4c = 64 = 16a + 4b + c = 16 4 To acquire rid of ‘c’ I will utilize this computation ; _16a + 4b + c = 16 4a + 2b + degree Celsiuss = 4 12a + 2b = 12 We can simplify this equation to: 6a + B = 6 My following computation is below: N =3