f (2) = a x 2³ + b x 2² + c x 2 + d
= 8a + 4b + 2c + d = 10 – equation 2
f (3) = a x 3³ + b x 3² + c x 3 + d
= 27a + 9b + 3c + d = 20 – equation 3
f (4) = a x 4³ + b x 4² + c x 4 + d
= 64a + 16b + 4c + d = 35 – equation 4
Equation 4 – Equation 3
I am doing this to eliminate d from the two equations, to create another equation.
64a + 16b + 4c + d = 35 – equation 4
- 27a + _ 9b + 3c + d = 20 – equation 3
37a + 7b + c = 15 – equation 5
Equation 3 – Equation 2
I am doing this to eliminate d from the two equations, to create another equation.
27a + 9b + 3c + d = 20 – equation 3
- 8a + 4b + 2c + d = 10 – equation 2
19a + 5b + c = 10 – equation 6
Equation 2 – Equation 1
I am doing this again to eliminate d from the two equations, to create another equations.
8a + 4b + 2c + d = 10 – equation 2
- a + b + c + d = 4 – equation 1
7a + 3b + c = 6 – equation 7
Equation 5 – Equation 6
I am doing this to eliminate c from the two equations, to make an equation.
37a + 7b + c = 15
- 19a + 5b + c = 10
18a + 2b = 5 – Equation 8
Equation 6 – Equation 7
I am doing this to eliminate c from the two equations, to create an equation.
19a + 5b + c = 10 – equation 6
- 7a + 3b + c = 6 – equation 7
12a + 2b = 4 – equation 9
Equation 8 – Equation 9
I am doing this to cancel out b and find the value of a.
18a + 2b = 5
12a + 2b = 4
6a = 1
a = 1/6
I am going to Substitute a = 1/6 into equation .
I’m doing this because to find what “b” is worth when substituting a = 1/6 into equation 9.
18a + 2b = 5
18 x 1/6 + 2b = 5
3 + 2b = 5
2b = 5 - 3
2b = 2
b = 2/2
b = 1
Substitute a = 1/6, b = 1 into equation 6.
I’m doing this because to find what “c” is worth when substituting a = 2, b = 3 into equation 6.
19a + 5b + c = 10
19 x 1/6 + 5 x 1 + c = 10
19/6 + 5 + c = 10
3 + 1/6 + 5 + c = 10
8 + 1/6 c = 10
c = 10 – 8 - 1/6
c = 11/6
I am going to Substitute a = 1/6, b = 1 and c = 11/6 into equation 1.
I’m doing this because to find what “d” is worth and also that leads me to working out the formula for the area of the triangle.
a + b + c + d = 4
1/6 + 1 + 11/6 + d = 4
12/6 + 1 + d = 4
2 + 1 d = 4
d = 4 –3
d = 1
I am going to Substitute a = 1/6, b = 1, c =11/6 and d = 1 put into the quadratic sequence.
F (n) = an³ + bn² + cn + d
F (n) = 1/6n³ + 1 x n² + 11/6n + 1
Now I have to try out the formula by putting in the n th term, prove that it works.
Substitute n=1
f(1) = 1/6 x 13 + 1 x 1² + 11/6 x 1 + 1
= 1/6 + 1 + 11/6 + 1
= 12/6 + 2
= 2 + 2
= 4
Substitute n=1
F(6) = 1/6 x 63 + 1 x 6² + 11/6 x 6 + 1
= 1/6 x 216 + 36 + 66/6 + 1
= 36 + 36 + 11 + 1
= 84
Substitute n=1
f (7) = 1/6 x 73 + 1 x 7² + 11/6 x 7 + 1
= 1/6 x 343 + 49 + 77/6 + 1
= 343/6 + 49 + 12 + 5/6 + 1
= 348/6 + 62
= 58 + 62
= 120
Term number (n) 1 2 3 4 5
Sequence 6 30 90 210 420
1st difference 24 60 120 210
2nd difference 36 60 90
3rd difference 24 30
4th difference 6
The formula for this would look like this
As the formula has 5 unknowns, I have to create 5 equations to find a, b, c, d and e.
f(1) = a x 14 + b x 13 + c x 12 + d x 1 + e = 6
= a + b + c + d + e = 6 – equation 1
f(2) = a x 24 + b x 23 + c x 22 + d x 2 + e = 30
= 16a + 8b + 4c + 2d + e = 30 – equation 2
f(3) = a x 34 + b x 33 + c x 32 + d x 3 + e = 90
= 81a + 27b + 4c + 3d + e = 90 – equation 3
f(4) = a x 44 + b x 43 + c x 42 + d x 4 + e = 120
= 256a + 64b + 16c + 4d + e = 120 – equation 4
f(5) = a x 54 + b x 53 + c x 52 + d x 5 + e = 210
= 625a + 125b + 25c + 5d + e = 120 – equation 5
As I have created 5 equations to find the Quartic sequence it is much difficult than the cubic sequence. I have use correct algebraic notation at right place and to take away each equation from another equation.
Equation 5 – Equation 4
I am doing this to eliminate e and to form a new equation which will be equation 6, when equation 5 – Equation 4.
625a + 125b + 25c + 5d + e = 420 - equation 5
- 256a + 64b + 16c + 4d + e = 210 - equation 4
369 + 61b + 9c + d = 210 - equation 6
Equation 4 – Equation 3
I am doing this to eliminate e and to form a new equation which will be Equation 7, when equation 4 – Equation 3.
256a + 64b + 16c + 4d + e = 210 - equation 4
- 81a + 27b + 9c + 3d + e = 90 - equation 3
175a + 37b + 7c + d = 120 - equation 7
Equation 3 – Equation 2
We are doing this to eliminate e and forming a new equation which will be Equation 8, when equation 3 – Equation 2.
81a + 27b + 9c + 3d + e = 90 – equation 3
- 16a + 8b + 4c + 2d + e = 30 – equation 4
65a + 19b + 5c + d = 60 - equation 8
Equation 2 – Equation 1
I am doing this to eliminate e and to create a new equation which will be equation 9, when equation 2 – equation 1.
16a + 8b + 4c + 2d + e = 30 – equation 2
a + b + c + d + e = 6 – equation 1
15a + 7b + 3c + d = 24 - equation 9
Equation 6 – Equation 7
I am doing this to eliminate d and to form a new equation which will be equation 10, when equation 6 – equation 7.
369a + 61b + 9c + d = 210 – equation 6
- 175a + 37b + 7c + d = 120 – equation 7
194a + 24b + 2c = 90 - equation 10
Equation 7 – Equation 8
I am doing this to eliminate d and to form a new equation which will be equation 11, when equation 7 – equation 8.
175a + 37b + 7c + d = 120 – equation 7
- 65a + 19b + 5c + d = 60 – equation 8
110a + 18b + 2c = 60 - equation 11
Equation 8 – Equation 9
I’m doing this because to eliminate d and forming a new equation which will be Equation 12, when equation 8 – Equation 9.
65a + 19b + 5c + d = 60
- 15a + 7b + 3c + d = 24
50a + 12b + 2c = 36 - Equation 12
Equation 10 – Equation 11
I am doing this to eliminate c and to create a new equation which will be equation 13, when equation 10 – equation 11.
194a + 24b + 2c = 90 – equation 10
- 110a + 18b + 2c = 60 – equation 11
84a + 6b = 30 - equation 13
Equation 13 – Equation 14
I am doing this to eliminate b and finding out what is the value of a.
84a + 6b = 30
- 60a + 6b = 24
24a = 6
a = 6/24
a = ¼
Substitute a = 1/4 into equation 13
I am going substitute the value of a= 1/4 into equation 13, so I could find the value of b.
84a + 6b = 30
84 x 1/4 + 6b = 30
84/4 + 6b = 30
21 + 6b = 30
6b = 30 – 21
6b = 9
b = 9/6
b = 3/2
Substitute a = ¼, b = 3/2 into equation 12.
I am going to substitute the value of a= 1/4 and b= 3/2 into equation 1, so I could find the value of c.
50a + 12b + 2c = 36
50 x ¼ + 12 x 3/2 +2c = 36
50/4 + 36/2 + 2c = 36
12½ + 18 + 2c = 36
30½ + 2c = 36
2c = 36 – 30½
2c = 5½
c = 5½ / 2
c = 2¾ or 11/6
Substitute a = ¼, b = 3/2, c = 11/4 into equation 9.
I am going to substitute the value of a=1/4, b= 3/2 and c= 11/4 into equation 9, so I could find the value of d.
15a + 7b + 3c + d = 24
15 x ¼ + 7 x 3/2 + 3 x 11/4 + d = 24
15/4 + 21/2 + 33/4 + d = 24
3¾ + 10½ + 8¼ + d = 24
22½ + d = 24
d = 24 – 22½
d = 1½ or 3/2
I am going to substitute the value of a= ¼, b=3/2, c= 11/4 and d= 3/2 into equation 1, I could find the value of e.
a + b + c + d + e = 6
¼ + 3/2 + 11/4 + 3/2 + e = 6
12/4 + 6/2 + e = 6
6 + e = 6
e = 6 – 6
e = 0
Substitute a = ¼, b = 3/2, c = 11/4, d = 3/2 and e = 0 into the general term of Quartic sequence.
f(n) = an4 + bn3 + cn2 + dn + e
Now this is the formula for the Quartic sequence.
Now I have to try out the formula by putting in the n th term, to prove that it works.
Substitute n = 6
F(n) = ¼n4 + 3/2n3 + 11/4n2 + 3/2n
F(6) = ¼ x (6)4 + 3/2 x (6)3 + 11/4 x (6)2 + 3/2 x 6
= ¼ x 1296 + 3/2 x 216 + 11/4 x 36 + 3/2 x 6
= 1296/4 + 648/2 + 396/4 + 18/2
= 324 + 324 + 99 + 9
= 648 + 108
= 756
Substitute n = 7
F(n) = ¼n4 + 3/2n3 + 11/4n2 + 3/2n
F(6) = ¼ x (7)4 + 3/2 x (7)3 + 11/4 x (7)2 + 3/2 x 7
= ¼ x 2401 + 3/2 x 343 + 11/4 x 49 + 3/2 x 7
= 2401/4 + 1029/2 + 539/4 + 21/2
= 735 + 525
= 1260
Factorization
Cubic sequence formula: -
This cubic has to be factorised, so first stage is to find the common number or letter. It should be put into brackets. When the formula is factorised, it should look like this.
I am doing this either see if there is any relationship between the factorised formulas or it is easier to plot the point in the graph.
I am going to find the value of the negative number so I could see how it goes in the graph.
F(n) = 1/6 (n³ + 6n² + 11n + 6) F(-1) = 1/6 ((-1)³ + 6 x (–1)² + 11 x -1 + 6)
F(0) = 1/6 (0³ + 6 x 0² + 11 x 0 + 6) = 1/6 ( -1 + 6 – 11 + 6)
= 1/6 (6) = 1/6 x 0
=1/6 x 6 = 0/6
= 6/6 = 1 = 0
F(-2) = 1/6 ((-2)³ + 6 x (–2)² + 11 x -2 + 6) F(-3) = 1/6 ((-3)³ + 6 x (–3)² + 11 x -3 + 6)
= 1/6 (-8 + 24 – 22 + 6) = 1/6 (-27 + 54 – 33 + 6)
= 1/6 x 0 = 1/6 x 0
= 0/6 = 0/6
= 0 = 0
F(- 4) = 1/6 ((-4)³ + 6 x (– 4)² + 11 x -4 + 6) F(-5) = 1/6 ((-5)³ + 6 x (-5)² + 11 x –5 + 6)
= 1/6 (- 64 + 96 – 44 + 6) = 1/6 (-125 + 150 – 55 + 6)
= 1/6 x –6 = 1/6 x -24
= - 6/6 = - 24/6
= -1 = - 4
F(-6) = 1/6 ((-6)³ + 6 x (– 6)² + 11 x -6 + 6)
= 1/6 (- 216 + 216 – 66 + 6)
= 1/6 x -60
= - 60/6
= -10
Using the results I have created a table which will be much easier to see and to plot the points in the graph“f (n) = 1/6 (n³ + 6n² + 11n + 6)”.
Quartic sequence formula: -
This Quartic has to be factorised, so first stage is to find the common number or letter. It should be put into brackets. When the formula is factorised, it should look like this. -
I am doing this either see if there is any relationship between the factorised formulas or it is easier to plot the point in the graph.
I am going to find the value of the negative number so I could see how it goes in the graph.
F(n) = 1/4n (n³ + 6n² + 11n + 6) F(-1) = 1/4n ((-1)³ + 6 x (–1)² + 11 x -1 + 6)
F(0) = 1/4n (0³ + 6 x 0² + 11 x 0 + 6) = 1/4n ( -1 + 6 + -11 + 6)
= ¼ x 0 x 6 = ¼ x –1 (0)
= 0 = - ¼ x 0 = 0
F(-2) = 1/4n ((-2)³ + 6 x (–2)² + 11 x -2 + 6) F(-3) = 1/4n ((-3)³ + 6 x (–3)² + 11 x -3 + 6)
= 1/4n ( -8 + 24 –22 + 6) = 1/4n ( -27 + 54 –33 + 6)
= ¼ x –2 x 0 = ¼ x –3 x 0
= 0 = 0
F(-4) = 1/4n ((-4)³ + 6 x (–4)² + 11 x -4 + 6) F(-5) = 1/4n ((-5)³ + 6 x (–5)² + 11 x -5 + 6)
= 1/4n (-64 + 96 – 44 + 6) = 1/4n (-125 + 150 –55 + 6)
= ¼ x – 4 x - 6 = ¼ x –5 x -24
= 24/4 = 6 = 120/4 = 30
F(-6) = 1/4n ((-6)³ + 6 x (–6)² + 11 x -6 + 6)
= 1/4n (-216 + 216 – 66 + 6)
= ¼ x – 6 x – 60
= 360/4 = 90
Using the results I have created a table which will be much easier to see and to plot the points in the graph “f(n) = 1/4n (n³ + 6n² + 11n + 6)”.
Now I have to draw a graph for both formulas:-
- 1/6 (n³ + 6n² + 11n + 6)
- 1/4n (n³ + 6n² + 11n + 6)
This graph is for the Cubic formula.
This graph is for the Quartic formula.
Conclusion
The relationship:
If you could see
-
1/6 (n³ + 6n² + 11n + 6)
-
1/4n (n³ + 6n² + 11n + 6)
that both of the formulas ends up with the same numbers and letters inside the brackets.
