Next Three Terms of a Quadratic Sequence

Warm-up:

(Care of )

The Problem:

Find the next three terms in this sequence

14, 24, 36, 50, 66, ..., ..., ...,

-/|\-

Solution:

Solution

As with any sequence we are given first let's look to see if it is not one we know. It's not one I know, so let's find the first difference between adjacent terms.

$n$	1		2		3		4		5
$n^{th}$ term $(U_{n})$	14		24		36		50		66
First difference $(U_{n + 1} - U_{n})$		10		12		14		16

This first difference is an arithmetic sequence with a common difference of 2, which enables us to work out the next three first differences:

$n$	1		2		3		4		5		6		7		8
$n^{th}$ term $(U_{n})$	14		24		36		50		66		?		?		?
First difference $(U_{n + 1} - U_{n})$		10		12		14		16		18		20		22

Adding these first differences in term to the previous term in the sequence we get:

$n$	1		2		3		4		5		6		7		8
$n^{th}$ term $(U_{n})$	14		24		36		50		66		84		104		126
First difference $(U_{n + 1} - U_{n})$		10		12		14		16		18		20		22

So the sequence is 14, 24, 36, 50, 66, 84, 104, 126,…

We can go further and find any term in the sequence. The first difference is an arithmetic sequence with a common difference of 2, implying that the original sequence is a quadratic sequence of the form $U_{n} = a n^{2} + b n + c$ , where a is the common difference divided by 2, ∴ $a = 1$ .

$n$	1		2		3		4		5		6		7		8
$n^{th}$ term $(U_{n})$	14		24		36		50		66		84		104		126
First difference $(U_{n + 1} - U_{n})$		10		12		14		16		18		20		22
$a n^{2}$	1		4		9		16		25		36		49		64
$U_{n} - a n^{2}$	13		20		27		34		41		48		55		62

But $U_{n} - a n^{2} = b n + c$ , so the last line is an arithmetic sequence with common difference $b = 7$ and zeroth term $c = 6$ . Therefore, the $n^{th}$ term of the sequence is given by $U_{n} = n^{2} + 7 n + 6$ .

-oOo-

Warm-up:

The Problem:

Solution:

Solution

Further Reading: