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• Level: GCSE
• Subject: Maths
• Word count: 3708

# Maths Coursework: Number Stairs

Extracts from this document...

Introduction

I have been set the task of working out the relationship between a 3 – step stair total and the position of the stair shape on the prearranged grid and other stair totals.

Aim:

As I work through this task I hope to find a formula that will help me to work out any stair total on a 10 by 10 grid.

Method:

I began by calculating the stair in a vertical sequence ranging from 1 step stair to 8 step stairs. The results of these calculations highlighted a definite pattern. These results were then summarised in a table from which a general formula was found.

Key:

= Base Number

n = Base number

1 Step Stair:

Difference =   +1                         +1                             +1

+1                                    +1                        +1

As we can see, the stair total for a 1-step stair using 1 as the base number on a 10 by 10-number grid is 1 as the total = 1. I can therefore conclude that the general equation for a 1-step stair on a 10 by 10 number grid is 1n + 0 = stair total where n is the base number. (Shaded box).

Total of step containing 1 as its base number – Difference

= 1- 1

= 0

## Formulae: 1n + 0 = Total

Testing the formula:

Total = 1n + 0

Total = 1 * 2 + 0

Total = 2 + 0

Total = 2

2 Step Stairs:

Difference = +3                          +3                                 +3

+3                              +3                                 +3

## 35

As we can see, the stair total for a 2-step stair using 1 as the base number on a 10 by 10-number grid is 14 as 1 + 2 + 11 = 14. I can therefore conclude that the general equation for a 2-step stair on a 10 by 10 number grid is 3n + 11 = stair total where n is the base number. (Shaded box).

Total of step containing 1 as its base number – Difference

= 14 – 3

= 11

Middle

Total = 36n + 924

Total = 36 * 2 + 924

Total = 72 + 924

Total = 996

## Total = 36n + 924

Final Formula:

Section 1:

= First Difference

= Second Difference

## 28

8

36

The data above is summarised in the table below.

 n 0 1 2 3 4 5 6 7 8 nth term 0 1 3 6 10 15 21 28 36 1st difference 1 2 3 4 5 6 7 8 2nd difference 1 1 1 1 1 1 1

Observation:

The differences between the nth terms were increasing which meant that I had to use quadratic sequences. The general term of a quadratic sequence is:

un =AN 2 + BN + C

a= the second difference in the sequence over 2.

###### b = is the coefficient of the term in n 2.

c = is the numerical value or constant value.

a = 2nd difference / 2

a = 1 / 2

Counting back, the zeros term is 0. So when n = 0, un = 0. When ‘n’ = 0, both ‘n 2’ and ‘bn’ are equal to 0 so ‘c’ = 0.

C = 0

To find the value of ‘b’, we look at the first term. When ‘n’ = 1, the value of u is 2.

un = n + bn

= 1 + b

2 = 1 + b

b = 1 / 2

1/2s2+ 1/2 s

Section 2:

= First Difference

= Second Difference

= Third Difference

## 616

8

924

The data above is summarized in the table below.

 n 0 1 2 3 4 5 6 7 8 nth term 0 0 11 44 110 220 385 616 924 1st difference 0 11 33 66 110 165 231 308 2nd difference 11 22 33 44 55 66 77 3rd difference 11 11 11 11 11 11

Nth term =an 3 + bn 2 + cn + d

When n = 1     13a+12b+1c+d=6                      Equation 1:  a + b + c + d = 0

When n = 2     23a+22b+2c+d=20                    Equation 2:  8a + 4b + 2c + d = 11

When n = 3     33a+32b+3c+d=52                    Equation 3:  27a + 9b + 3c + d = 44

When n = 4     43a+42b+4c+d=108                  Equation 4:  64a + 16b + 4c + d = 110

Subtracting equation 1 from each of the other 3 gives:

Equation 5           7a+3b+c=11

Equation 6           26a+8b+2c=44

Equation 7           63a+15b+3c=110

Subtracting 2 times equations 5 from equation 6 gives:

Conclusion

an3 + bn2 + cn + d:

##### If n = 1 , 2 , 3 , 4 , 5

a+b+c+d , 8a+4b+2c+d , 27a+9b+3c+d , 64a+164+4c+d , 125a+25b+5c+d , 7a+3b+c , 9a+5b+c , 37a+7b+c , 61a+9b+c, 12a+2b , 18a+2b , 24a+2b

6a , 6a

I am now going to match the numbers in the cubic sequence formula above to the numbers that correspond to their positions in the co-efficient in front of g sequence. 6a =

a = 1/6

12a+2b = 3

12(1/6) + 2b = 3

2 + 2b = 3

2b = 1

b = 0.5

7a+3b+c= 3

7(1/6)+3(1/2)+c = 3

c= 1/3

a+ b + c + d = 1

1/6 + ½ + 21/3 + d = 1.

d = 0

Therefore the equation for this part of the sequence will be :

(1/6 n3 + 1/2n2 + 1/3n) = coefficient of g

Section 3: The Sequence For The Constant

The sequence for the constant will go exactly the same as the sequence for the co-efficient of g as all these values are the same. Therefore the equation for this part of the sequence will be :

(1/6 n3 + 1/2n2 + 1/3n) = the constant

Conclusion

##### Now that we have finished the three parts of the sequence we can put them together in order to make one equation for any size step-stair on any size grid.

(1/2 n2 + 3/2n + 1) x + (1/6 n3 + 1/2n2 + 1/3 n) g + (1/6 n3 + 1/2n2 + 1/3 n) = stair total

x means the number of squares in the stair shape..

g stands for grid size.

n is the equation number. (If you don't know the equation number it is the same number as the number of squares in the 2nd column from the left of the stair shape).

Reference:

• Internet: www.y-maths.co.uk/index.htm
• Higher GCSE for Edexcel
• Edexcel GCSE Mathematics higher course
• GCSE Mathematics Revision Guide Higher Level

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