Quadratic sequences
At least four numbers are needed to determine whether the sequence is quadratic or not.
Consider this number pattern 6, 12, 22, 36, 54.
There is no common difference between the numbers.
The differences are 6, 10, 14, 18.
Now we can see if there is a second common difference.

In this sequence, there is a second common difference of 4. The next term will be:
a₆=54+(18+4)=76

A pattern with a common second difference is called a quadratic number sequence.

The general formula for any term of a quadratic sequence is: a_n=an²+bn+c

If a_n=an²+bn+c then 2a is the second difference,

a+b is a₂–a₁,
a+b+c is the first term.

Example 1. Look at the number sequence 12, 20, 32, 48, …

Second common difference is 4. So 2a=4 ∴ a=2.
② a₂–a₁=8. So 3a+b=8 ∴3⋅2+b=8∴b=2
③ 1st term is 12. So a+b+c=12∴2+2+c=12∴c=8
∴a_n=2n²+2n+8
a₅=2⋅5²+2⋅5+8=68
a₆=2⋅6²+2⋅6+8=92

Activity 1.
1. Consider the number pattern: 3, 13, 31, 57, 91, …
a) Determine the general term for this pattern.
b) Calculate the 7th term of this pattern.
c) Which term is equal to 241?

2. Find term 6 of this pattern and then find the rule in the form a_n=an²+bn+c.

-1, 3, 9, 17, 27, …
Solutions
1. a) It helps to make a diagram:

∴ it is a quadratic sequence.
2a=8 ∴a=4
3a+b=10 ∴3(4)+b=10
b=-2
a+b+c=3 ∴4+(-2)+c=3
c=1
∴a_n=4n²-2n+1
b) a₇=4⋅7²-2⋅7+1
=4⋅49-14+1=183
c) 241=4n²-2n+1
Make the equation =0 to solve.
0=4n²-2n+1-241
0=4n²-2n-240
Divide through by 2.
0=2n²–n-120
0=(2n+15)(n-8)
factorise
2n+15=0 ∴n=-7.5 or
n-8=0 ∴n=8
n=-75 not possible because n is tne position of the term so it must be a positive natural number.
∴ 241 is the 8th term of the sequence.

2)

Use the pattern of the numbers.
a₆=27+(10+2)+39
2a=2 ∴a=1
3a+b=4
3(1)+b=4 ∴b=1
a+b+c=-1
1+1+c=-1 ∴c=-3
a_n=n²+n-3
Check a₆.
a₆=6²+6-3=39