- Subject: Mathematics
- AS: 91028
- Level: 1
- Credits: 4
Mathematics 1.3 Investigate relationships between tables, equations and graphs
e.g. Find a rule for calculating the total number of boxes in each diagram, if the number of boxes in the bottom layer is known.
Draw up a table to show the number of boxes in the bottom layer and the total number of boxes in each diagram.
Calculate the differences between the terms.
|No. of boxes in bottom layer (n)||1||2||3||4||5|
|Total no. of boxes (t)||1||3||6||10|
First difference 2 3 4 Second difference 1 1
Let the number of boxes in the bottom layer be n and the total number of boxes be t.
Because the second differences are the constant (they are all equal to 1), the rule in this example will be a quadratic expression.
The rule will be in the form: t = an² + bn + c – where a, b, and c are constants to be found.
Now when n=1, t=1 so 1 = a + b + c -------- (1)
When n=2, t=3 so 3 = 4a + 2b + c -------- (2)
When n=3, t=6 so 6 = 9a + 3b + c -------- (3)
Subtract equation (1) from equation (2) to eliminate c:
2 = 3a + b -------- (4)
Subtract equation (2) from equation (3) to eliminate c:
3 = 5a + b -------- (5)
This now gives two equations in a and b.
Solve these simultaneously to find a and b.
Subtract equation (4) from equation (5):
1 = 2a
Substitute this value for a in equation (4) to find b =
Substitute these values for a and b into equation (1) to find c = 0.
Write a = , b = and c = 0 into the rule.
Hence the rule for this pattern is t = n² + n
This method may be shortened by remembering that
a+b+c = first term
3a+b = first difference
2a = second difference
For example for the pattern 10, 18, 28,.....
The equation goes an² + bn + c
10 18 28
a+b+c = 10
3a+b = 8
2a = 2
Therefore a = 1, b = 5 and c = 4. Equation is n²+5n+4