Quadratic Sequences

What makes a sequence quadratic?

A sequence is quadratic when its $n$th term can be written in the form

$an^2+bn+c,$

where $a$, $b$ and $c$ are constants and $a\ne0$. The highest power of the term number $n$ is therefore 2.

Quadratic sequences do not have a constant . Instead, their are constant. Consider

$3,\ 8,\ 15,\ 24,\ldots$

Subtract each term from the next:

The first differences are not equal, so this is not an arithmetic sequence. Their differences are both 2, so it has the pattern of a quadratic sequence.

In a GCSE sequence question, constant non-zero second differences identify the intended quadratic pattern. Constant first differences instead indicate an arithmetic sequence.

Why the second differences are constant

The square numbers begin

$1,\ 4,\ 9,\ 16,\ 25,\ldots$

Their first differences are $3,5,7,9,\ldots$, which increase by 2 each time. Therefore, the second difference of $n^2$ is 2.

Multiplying $n^2$ by $a$ multiplies every difference by $a$. The second difference of $an^2$ is consequently $2a$. Adding a linear or a constant does not change the second differences. This gives the useful connection

$a=\frac{\text{constant second difference}}{2}.$

For example, a second difference of 6 gives $a=3$, while a second difference of $-4$ gives $a=-2$.

Continuing a quadratic sequence

Constant second differences let you extend a sequence even though there is no single number to add repeatedly.

Take

$2,\ 8,\ 18,\ 32,\ 50,\ldots$

The first differences are $6,10,14,18$, and the second differences are all 4. The next first difference must therefore be

$18+4=22.$

Add this new first difference to the last term:

$50+22=72.$

So the next term is 72. To continue again, increase the first difference by another 4: the following difference is 26 and the following term is $72+26=98$.

This order matters: extend the constant second differences, use them to extend the first differences, and then use those to extend the original terms.

Finding the $n$th term

Return to

$3,\ 8,\ 15,\ 24,\ldots$

Its constant second difference is 2, so the coefficient of $n^2$ is

$a=\frac{2}{2}=1.$

Begin with $n^2$ and compare its values with the sequence:

The remaining sequence $2,4,6,8,\ldots$ has $n$th term $2n$. Combining the quadratic and linear parts gives

$u_n=n^2+2n.$

Check the rule: when $n=1$, it gives $1^2+2(1)=3$; when $n=4$, it gives .

This comparison method works generally:

1. Calculate the first and second differences.
2. Divide the constant second difference by 2 to find $a$.
3. Generate the values of $an^2$ for the required positions.
4. Subtract these from the given terms.
5. Find the linear rule $bn+c$ for the remaining sequence.
6. Combine the parts to obtain $an^2+bn+c$ and check it.

When the quadratic coefficient is negative

The same method applies when the second differences are negative. For

$-17,\ -30,\ -49,\ -74,\ -105,\ldots$

the first differences are $-13,-19,-25,-31$, so the constant second difference is $-6$. Therefore,

$a=\frac{-6}{2}=-3.$

The values of $-3n^2$ are $-3,-12,-27,-48,-75$. Subtracting these from the given terms leaves

$-14,\ -18,\ -22,\ -26,\ -30,$

whose $n$th term is $-4n-10$. The complete rule is therefore

$u_n=-3n^2-4n-10.$

A negative second difference must give a negative coefficient of $n^2$.

Using a quadratic $n$th term

An $n$th-term rule gives any term directly. If

$u_n=3n^2-10,$

then the fifth term is found by substituting $n=5$:

$u_5=3(5^2)-10=3(25)-10=65.$

Square the term number before carrying out the multiplication and addition or subtraction. The notation $3n^2$ means $3\times n^2$, not $(3n)^2$.

Recognise

For a quadratic sequence: