5 13 25 41 25 13 5 n = 5
My table of results shows that two identical layers are added each time and also, as you look down each column you can see the progression of each individual layer. As I mentioned earlier, it progresses the same as the 2D patterns, which is how I worked out the number of cubes in the each layer of pattern 5. This is what the bottom line of my pyramid table looks like:
I can also work out the number of cubes in the centre layer algebraically by using the formula:
Nth term = 2n – 2n + 1
(that being the formula for the 2D pattern)
In order to work out the number of cubes in the centre layer I must use this formula with n being = 5:
2n – 2n + 1
= (2x5x5) – (2x5) + 1
= 50 – 10 + 1
= 41 cubes.
Total Number of Cubes in Each Pattern
Now that I know how many cubes each layer contains I could either draw pattern 5, or simply add the bottom line of my pyramid table.
After drawing this shape and assuming that it is in 3D I can see that it has a total of 129 cubes,
1 + 5 + 13 + 25 + 41 + 25 + 13 + 5 + 1 = 129 cubes.
I needed to work out the number of cubes in pattern 5 in order to find the nth term of the sequence using algebra.
Pattern No. n = 1 2 3 4 5
Sequence = 1 7 25 63 129
1st diff. = 6 18 38 66
2nd diff = 12 20 28
3rd diff = 8 8
In order to find the nth term I will use the general cubic polynomial:
an + bn + cn + d
I can use this formula to help me work out the nth term by changing n into a number. For instance n=2, so an equation could be:
2nd term = a x 2 + b x 2 + c x 2 + d
= 8a + 4b + 2c + d
To begin solving the equation, I substituted the numbers in the polynomial for all the values of n:
n = 1 1st term = a + b + c + d = 1……………………..1)
n = 2 2nd term = 8a + 4b + 2c + d = 7………………...2)
n = 3 3rd term = 27a + 9b + 3c + d = 25……………....3)
n = 4 4th term = 64a + 16b + 4c + d = 63……………..4)
n = 5 5th term = 125a + 25b + 5c + d = 129…………..5)
The blue numbers were gained from the pattern sequence so I have done the first step in trying to find the nth term. I am now going to use differencing with algebra to find the nth term.
Equation 2) – 1) 8a + 4b + 2c + d = 7
a + b + c + d = 1
7a + 3b + c = 6………………A)
Equation 3) – 2) 27a + 9b + 3c + d = 25
8a + 4b + 2c + d = 7
19a + 5b + c = 18……………B)
Equation 4) – 3) 64a + 16b + 4c + d = 63
27a + 9b + 3c + d = 25
37a + 7b + c = 38……………C)
Equation 5) – 4) 125a + 25b + 5c + d = 129
64a + 16b + 4c + d = 63
61a + 25b + 5c + d…………..D)
I have labelled the equations that I have just found A-D (coloured green) and I am about to use them in the third step of working out the nth term.
These are the second group of differences:
Equation B) – A) 19a + 5b + c = 18
7a + 3b + c = 6
12a + 2b = 12 (this can be simplified to 6a+b=6)…..E)
Equation C) – B) 37a + 7b + c = 38
19a + 5b + c = 18
18a + 12b = 20 (this can be simplified to 9a+b=10)..F)
Equation D) – C) 61a + 25b + 5c + d
37a + 7b + c = 38
24a + 2b = 28 (this can be simplified to 12a+b=14)..G)
The equations that are green are the ones I will use for the fourth step in finding the nth term. I can now work out the 3rd group of differences:
Equation F) – E) 9a + b = 10
6a + b = 6
3a = 4 (this can be simplified to a = 4/3)
Equation G) – F) 12a + b = 14
9a + b = 10
3a = 4 (this can be simplified to a = 4/3)
I now have a number for ‘a’ in the cubic polynomial and now if I use equation 1) I can find ‘b’.
(6 x 4/3) + b = 6
8 + b = 6
b = -2
Now if I put ‘a’ and ‘b’ into equation A I will be able to find the value of C.
(7 x 4/3) + (3 x -2) + c = 6
28/3 – 6 + c = 6
28/3 – 18/3 + c = 18/3
c=8/3
By putting ‘a’, ‘b’ and ‘c’ into equation 1. I can work out the value of ‘d’
4/3 – 2 + 8/3 + d = 1
4/3 – 6/3 + 8/3 + d = 1
6/3 + d = 1
2 + d = 1
d = -1
Now that I have the values for ‘a’, ‘b’, ‘c’ and ‘d’ I can substitute them into the polynomial:
Nth term = 4/3n - 2n + 8/3n – 1
Now I need to check the formula to make sure it is correct. I will use n=2.
N=2
Second term = (4/3 x 2 ) – (2 x 2 ) + (8/3 x 2) – 1
= 32/3 – 8 + 16/3 – 1
= 48/3 – 9
= 16 – 9
= 7
As the total number of cubes in pattern 2 is 7 this formula appears to be correct.
