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• Level: GCSE
• Subject: Maths
• Word count: 1612

rectangles. I will be trying to develop a formula that will enable me to calculate the sum of all the numbers in a rectangle given

Extracts from this document...

Introduction

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The value of the rectangle is 132.

We would get the value by calculating the sum of all the numbers.

A way we could find this value out could be by using a formula. But what formula could we use?

 n n+1 n+2 n+10 n+11 n+12
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The sum of the values in the turquoise box adds up to 132.

To find this, we could try n and n+12.

16+16+12=44

132÷44=3

3 are equal to the width.

So far we have w (2n + 12).

Middle

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This time, the width of each shaded square is 4.

For the first shaded square, the values come up to 180.

4(2n+12): 4(44): 4x44= 176.

As you can see, we have changed the width, but have not changed the n+12 to n+13. If we tried this, we would get 180.

To do this so our formula is correct we would have to alter it.

W (2n+r). ‘R’ stands for the range. To make sure this formula is correct, I will try it on the remaining two squares and then on one with a completely different width.

The sum of the rose coloured box comes up to 380.

Conclusion

R(LxW)+(LxW)

2

14(2x5)+(2x5)

2

14(10): 140. 140÷2=70. 70+(10)=80. So far this formula has been correct, but would happen if initial term changes from 1 to a different number.

R(LxW)+(LxW)

2

12(2x3)+(2x3)

2

12(6)=72. 72÷2=36. 36+6=42.

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I could now try to split the height from 2 to 1 and adding both rows to get the final product.

w(in+1). I can change in which is the initial term to n, which will stand for the lowest number.

Width(n+1). 3(4+1)=3(5)=15. 4+5+6=15.

For the second row we will have 3(14+1)=3(15)= 45. 14+15+16=45.

45+15+60.

4+5+6+14+15+16=60.

54+55+56+57+64+65+66+67=484.

4(54+1) 4(55)= 220.

4(64+1) 4(65) 195. 220+260=480. This answer is incorrect. So instead of it being w(n+1), I am going to change it to w(n+(r-1)) for each row.

4 (54+(2))= 4(56)= 224

4(64+2)=4(66)=264. 224+264= 488. this answer is also incorrect.

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