Quadratic Sequences

From positions to terms

In a sequence, $n$ represents the position: $n=1$ for the first term, $n=2$ for the second, and so on. An nth-term formula lets you calculate a term directly from its position.

A quadratic sequence has an nth term of the form

$u_n=an^2+bn+c,$

where $a$, $b$ and $c$ are constants and $a\ne0$. The square numbers illustrate the simplest case:

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

Their rule is $u_n=n^2$.

Sometimes a sequence can be recognised by comparison with the square numbers. For example,

$2,\ 5,\ 10,\ 17,\ldots$

is one more than $1,4,9,16,\ldots$, so its nth term is $n^2+1$.

How differences reveal a quadratic rule

For a linear sequence, the first differences are constant. In a quadratic sequence, the first differences change, but their differences—the second differences—are constant.

Consider

$-1,\ 1,\ 7,\ 17,\ 31,\ldots$

The constant second difference shows that a quadratic rule is appropriate.

There is also an important connection between the second difference and the coefficient of $n^2$. If

$u_n=an^2+bn+c,$

then the difference between consecutive terms is

$u_{n+1}-u_n=2an+a+b.$

This first difference increases by $2a$ whenever $n$ increases by 1. Therefore:

$\text{constant second difference}=2a,$

so

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

Finding the complete nth term

Use the sequence $-1,1,7,17,31,\ldots$ again. Its second difference is $4$, so

$a=4\div2=2.$

The rule must therefore begin with $2n^2$:

$u_n=2n^2+bn+c.$

Now calculate the values of $2n^2$ and subtract them from the original terms.

Removing the quadratic part leaves the arithmetic sequence

$-3,\ -7,\ -11,\ -15,\ -19,\ldots$

Its common difference is $-4$, so its nth term has the form $-4n+c$. Substituting $n=1$ gives

$-4(1)+c=-3,$

so $c=1$. The remainder is therefore $-4n+1$. Add this to the quadratic part:

$\boxed{u_n=2n^2-4n+1}.$

This gives the general method:

1. Find the first differences.
2. Find the second differences and check that they are constant.
3. Divide the constant second difference by 2 to obtain $a$.
4. Write the terms of $an^2$ beneath the original terms.
5. Calculate original term $-an^2$ at every position.
6. Find the linear nth term of the remaining arithmetic sequence.
7. Combine the two parts to obtain $an^2+bn+c$.

Checking the formula

Substitute several position numbers into the proposed rule. For $u_n=2n^2-4n+1$:

• $u_1=2(1)^2-4(1)+1=-1$
• $u_2=2(2)^2-4(2)+1=1$
• $u_3=2(3)^2-4(3)+1=7$

These reproduce the given terms. Checking several positions is important because an arithmetic slip while finding the differences can produce a plausible-looking but incorrect rule.

Applying the same method to the OCR example

For

$0,\ 3,\ 10,\ 21,\ldots,$

the first differences are $3,7,11$, so the second differences are $4,4$. Hence $a=2$. Subtracting $2n^2$, whose terms are $2,8,18,32$, leaves

$-2,\ -5,\ -8,\ -11,\ldots$

This remainder has nth term $-3n+1$. Therefore

$\boxed{u_n=2n^2-3n+1}.$

For example, at $n=4$ the formula gives $2(4)^2-3(4)+1=21$, matching the fourth term.

Recognise the structure

A quadratic sequence has

$u_n=an^2+bn+c.$

Its first differences are not constant, but its second differences are constant.

Find the formula

1. Calculate first and second differences.
2. Use $a=\frac{\text{constant second difference}}{2}.$
3. Subtract the sequence $an^2$ from the original terms.
4. Find the linear nth term $bn+c$ of the remainder.
5. Combine the parts: $u_n=an^2+bn+c$.

Check

Substitute $n=1,2,3$ into the formula and confirm that it reproduces the corresponding terms. Remember that the first term normally corresponds to $n=1$.