PROBLEM SOLVING WITH PATTERNS USING QUADRATIC FUNCTON: f(n) = an2 + bn + c
nth term
:
sequence f(n) :
1
5
first difference:
2
14
9
second difference:
3
27
13
4
4
44
17
4
5
65
21
4
Note: when a common difference is found on the second difference, the sequence is quadratic
and the general formula for this model is f(n) = an2 + bn + c
nth term (n)
f(n)
1
5
f(n1) = 5
2
14
f(n2) = 14
3
27
f(n3) = 27
4
44
5
65
Use the following formula to solve for a, b and c using systems of linear equations
f(n1) = a + b + c

5=a+b+c
 equation 1
f(n2) = 4a + 2b + c

14 = 4a + 2b + c
 equation 2
f(n3) = 9a + 3b + c

27 = 9a + 3b + c
 equation 3
Eliminate c using equation 2 and equation 1
14 = 4a + 2b + c
– (5 = a + b + c)
------------------------9 = 3a + b  equation 4
Eliminate c using equation 3 and equation
27 = 9a + 3b + c
– (5 = a + b + c)
---------------------22 = 8a + 2b
11 = 4a + b  equation 5
Eliminate b and solve for a using equations (4) and (5)
11 = 4a + b  equation 5
– (9 = 3a + b)  equation 4
-------------------------------2 =a
Solve for b using either equations (4) or (5)
Using equation 4:
9 = 3a + b
9 = 3(2) + b
9=6+b
9–6=b
3=b
Solve for c using either equations (1), (2) or (3)
Using equation 1:
5=a+b+c
5=2+3+c
5=5+c
5–5=c
0=c
Since a = 2, b = 3 and c = 0, thus the mathematical model is: f(n) = 2n2 + 3n or f(n) = n(2n + 3)
To test:
let n = 3
f(n) = 2n2 + 3n
f(n) = 2(3)2 + 3(3)
f(n) = 2(9) + 9
f(n) = 27
let n = 5
f(n) = 2n2 + 3n
f(n) = 2(5)2 + 3(5)
f(n) = 2(25) + 15
f(n) = 65
Example 2: On the first week, Paul saved ₱4 of his weekly allowance. On the next week, he saved ₱7 then ₱12
and ₱19 on the 3rd and 4th weeks, respectively. How much will he save on the 12th and 20th weeks?
n
f(n)
1
4
2
7
3
3
12
5
2
n
f(n)
1
4
f(n1) = 4
2
7
f(n2) = 7
4
19
…
…
12
?
…
…
20
?
7
2
(second-degree function)
3
12
f(n3) = 12
4
19
…
…
12
?
…
…
20
?
Use the following formula to solve for a, b and c using systems of linear equations
f(n1) = a + b + c

4=a+b+c
 equation 1
f(n2) = 4a + 2b + c

7 = 4a + 2b + c
 equation 2
f(n3) = 9a + 3b + c

12 = 9a + 3b + c
 equation 3
Eliminate c using equation 2 and equation 1
7 = 4a + 2b + c
– (4 = a + b + c)
------------------------3 = 3a + b  equation 4
Eliminate c using equation 3 and equation
12 = 9a + 3b + c
– (4 = a + b + c)
---------------------8 = 8a + 2b
4 = 4a + b  equation 5
Eliminate b and solve for a using equations (4) and (5)
4 = 4a + b  equation 5
– (3 = 3a + b)  equation 4
-------------------------------1=a
Solve for b using either equations (4) or (5)
Using equation 4:
3 = 3a + b
3 = 3(1) + b
3=3+b
3–3=b
0=b
Solve for c using either equations (1), (2) or (3)
Using equation 1:
4=a+b+c
4=1+0+c
4=1+c
4–1=c
3=c
Since a = 1, b = 0 and c = 3, thus the mathematical model is: f(n) = (1)n2 + (0)n + 3  f(n) = n2 + 3
To test the model:
let n = 3
f(3) = (3)2 + 3
f(n) = 9 + 3
f(n) = 12
n=4
f(4) = (4)2 + 3
f(4) = 16 + 3
f(4) = 19