a=4, b=0. Now I have the equation, 4n+0, now let’s test it! We know the first terms is 4, so 4xn(1)+0, 4x1=4 so the answer is 4 and that is the 1st term.

Now ill do the same with the red squares

To find this equation I will need different bits of information, the equation is an2+bn+c,

A is ½ of the second difference, in this case a = 2

B is complicated, to find b you will need to find c first and substitute a and c into the equation

C is the 0th term, in this case 1, I will also need an n (term) I will choose the 1st term which is 1, now to find the equation

1)Un=an²+bn+c

2)1= an²+bn+c

3)1=2×1²+b×1+1

4)1=2+b+1

5)1=3+b

6)1-3=2+b-3

7)-2=b

Now I have the full equation and now I know b=-2 so I have the equation

n=4n²+-2n+0, we’ll test it with term 1, term 1=1, 2×1²+-2×1+1 which equals 1, hey it works!

Now to find the equation for the total squares, because you have seen be do two equation, you must understand them, so now let’s do the next one without me explaining it to you, 1st step find the differences

2nd Step find the equation, a=2 c=1 now to find b

1)Un=an²+bn+c

2)5= an²+bn+c

3)5=2×1²+b×1+1

4)5=2+b+1

5)5=3+b

6)5-3=3+b-3

7)2=b

Let’s test it; term 1 is 5, 2×1²+2×1+1=5. I’m right again! Now we have the equations for the black, white and total squares

Black Squares= 4n+0

Red Squares = n=2n²+-2n+1

Total Squares = n=2n²+2n+1

Borders: Part 2

Now to expand my investigation into the 3rd dimension. Please find the sheet with my drawing of the 3D shapes on, after you have seen that continue reading

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Here is where maths becomes complicated, the drawings on the last page are the cross shaped in the 3rd dimension, and here is the table showing how many cubes each shape has.

Now I will find the differences

The cubic equation is more complicated than the 2 I have been through, cubic expression is [an³+bn²+cn+d] a is factorial 3 and to find it I will need to divide the 3rd difference by factorial 3 (3!) Factorial three equals 1×2×3 which equals 6, so basically a is just the 3rd difference divided by 6, in this case the 3rd difference is 8, to find be you take away an³ from an³+bn²+cn+d which leaves you with a quadratic equation, here is how you do it

Here we go now were left with the quadratic equation, now I will find its differences.

Now we can find b, c and d. b, c and d are basically the a, b and c in the quadratic equation, in the same way we have to find c last and b and c first, b=½ of the second difference, in this case 2, d= 0th term, in this case 1, now we substitute b and d into the equation to find c:

1)Un=bn²+cn+d

2)27= an²+cn+c

3)27=2×3²+3c×3+1

4)27=18+3c+1

5)27=19+3c

6)27-19=19+3b-19

7)8=3b

8)

9)2=c

Now we have c, the full equation is n³+2n²+n+1. Our journey doesn’t end here, we still need to test it, let’s go ahead. We will test it with the 1st term, the 1st term is 7, ×1³+2×1²+×1+1=7 Hooray, the equation works.