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# GCSE: Consecutive Numbers

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Meet our team of inspirational teachers Get help from 80+ teachers and hundreds of thousands of student written documents 1. ## GCSE Maths questions

• Develop your confidence and skills in GCSE Maths using our free interactive questions with teacher feedback to guide you at every stage.
• Level: GCSE
• Questions: 75
2. ## The Towers of Hanoi is an ancient mathematical game. The aim of this coursework is to try to identify patterns and rules associated with the game and explain them in mathematical terms.

It is clear that there is an element of doubling involved, as the least number of moves nearly doubles each time. When I add the extra column see above, it is clear that there is a doubling element involved. When I look again, I can see that the pattern is the previous term doubled plus 1. This can be expressed mathematically as: Un = 2(Un-1) +1 This can be shown in: 1. For 1 disc, it takes 1 move to move disc A from pole 1 to pole 3; 2.

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3. ## Investigate calendars, and look for any patterns.

11 Monday 12 Wednesday Therefore, we can see that the dates are thus: 1, 4, 7 = same 2, 8 = same 3, 11 = same 5 6 9, 12 = same 10 As you can see, the dates which fall on the same day are not the same as in the study sample Ex 1.1, so there is no pattern here. However, to make this a fair test I will check another year, 2003, to be sure of this, as it could be an anomalous result of some kind.

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4. So N=N(s,t), or s=s(N) and t=t(N), where . Here I list the s and t parameters for the first five triangular squares: 1. s=1, t=1. t/s=1. 36. s=6, t=8. t/s=1.333... 1225. s=35, t=49. t/s=1.4. 41616. s=204, t=288. t/s=1.411764705... 1413721. s=1189, t=1681. t/s=1.413793103... I also listed the ratio t(N)/s(N) for each of these N. Notice how the t/s ratio approaches as N gets bigger. Also notice that t is either a perfect square or a perfect square minus one, viz.

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5. ## Transforming numbers

To do a thorough investigation, I decided to use excel spreadsheet with a > b, by 2,3,4,and so on. It always gave the same result, transformation converging towards V2. This led me to my next step to investigate what happens when a is less than b. Table3 a=5, b=7 a less than b by 2 (a<b) Sequence Sequence of of Numerator Denominator Result 5 7 0.7142857 19 12 1.5833333 43 31 1.3870968 105 74 1.4189189 253 432 1.4143519 It is seen from table 3 that, even when a is less than b, it still converges to V2.

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6. ## Investigation to Find the number of diagonal of any 2 Dimensional or / and 3 Dimensional - A diagonal is a line drawn from one vertex (corner) of the shape to another

In this experiment I am going to require the following: A calculator A pencil A pen Variety of sources of information Paper Ruler In this investigation I have been asked to find out how many squares would be needed to make up a certain pattern according to its sequence. The pattern is shown on the front page. In this investigation I hope to find a formula which could be used to find out the number of squares needed to build the pattern at any sequential position.

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7. ## Beyond Pythagoras

Perimeter = shortest +middle + longest length 5 + 12 + 13 = 30 units 7 + 24 + 25 = 56 units I found the area for the sequences 5, 12, 13 and 7, 24, 25, by simply finding the product (multiply) the shortest and middle and halving the answer. E.g. Area = 1/2 x shortest x middle length 1/2 x 5 x 12 = 30 square units 1/2 x 7x 24 = 84 square units I used this method for the area because it is a right-angled triangle.

• Word count: 2095
8. ## Nth Term Investigation

The nth for this one is (n-1)2. Here are my predictions for other squares with different lengths. n x n + 10 x 10 4 36 81 25 x 25 4 96 576 50 x 50 4 196 2401 100 x 100 4 396 9801 Rectangles No.1 I will now move on to rectangles, as there are so many ways of doing rectangles I will start with one side going up by one each time (x) and the other side always being the same (t) which will be 2.

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9. ## Consecutive Numbers Investigation

I will now hope to show that it works with algebra. X, X+1, X+2 1st *3rd = X*(x+2) = X2=2X 2nd squared = (X+1)2 = (X+1)(X+1) (X +1)(X+1) = X2 + 1 + 1X + 1X = X2 + 2X + 1 The only difference is +1. It shows that the difference will always be 1. I am now going to see what happens if I make the gap 2. Gap 2 3, 5, 7 3*7 = 21 52 = 5x5 = 25 Difference 4 5, 7, 9 5*9 = 45 72 = 7*7 = 49 Difference 4 17, 19, 21 17*21 = 357 192 = 19*19 = 361 Difference 4 It would appear that it would work every time.

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