# On quadratic sequences

*This is a guest post from Mark Ritchings, a maths tutor in Bury.*

A quadratic sequence is a sequence for which the $n$th term is $an^2+bn+c$. The constants $b$ and/or $c$ might be zero but $a$ definitely isn’t.

- The first term is $a\times 1^2 +b \times 1 + c = a+b+c$.
- The second term is $a \times 2^2 +b \times 2 + c = 4a+2b+c$.
- The third term is $a \times 3^2+b\times 3+c=9a+3b+c$.

… so, if you know the values of $a$, $b$ and $c$ you can work any term of the sequence.

In their wisdom, the people who decide what mathematics you have to learn and what you will be tested on have decided that you need to be able to do at least the following things:

- Given the expression for the $n$th term of a sequence, work out the first few terms, or the 20th term, or any given term.
- Given a sequence, work out the next few terms or figure out the $n$thterm.

I’m going to focus on the second of these points. Here’s an example:

If the first four terms of a quadratic sequence are

6, 17, 34, and 57, find the next term and the $n$th term.

In the table below the “working out” has been added below. The first differences are found by selecting each term, after the first, and subtracting the previous term. The second difference is found in a similar way. I hope that the pattern which allows you to work out the next term is clear. The next term is 86.

Sequence | 6 | 17 | 34 | 57 | 86 | ||||

1st difference | 11 | 17 | 23 | 29 | |||||

2nd difference | 6 | 6 | 6 |

There are several ways to find the $n$th term of a sequence.

### The usual method

The most popular method with teachers seems to be the following:

- Find half of the second difference. For the example above, this is 3.
- This is the coefficient of $n^2$ in the required $n$th term, because your teacher says so. ((Zing! - Ed.))
- Work out the first few terms of the sequence for which the $n$th term is $3n^2$. These would be $3 \times 1^2 = 3$, $3\times2^2=12$, and $3 \times 3^2 = 27$ - so that sequence begins 3, 12, 27.
- Subtract these numbers from the terms of the given sequence to obtain a linear sequence.

$3n^2+?n+?$ | 6 | 17 | 34 |

$3n^2$ | 3 | 12 | 27 |

$?n+?$ | 3 | 5 | 7 |

In this example the linear sequence is 3, 5, 7, …

Write down the $n$th term of this linear sequence. This is $2n+1$. The question marks in the table above can be replaced.

$3n^2+2n+1$ | 6 | 17 | 34 |

$3n^2$ | 3 | 12 | 27 |

$2n+1$ | 3 | 5 | 7 |

The required $n$th term is $3n^2+2n+1$.

But why is the coefficient of $n^2$ half the second difference? Looking at the general case answers this question. In the table below you can see that the second difference is $2a$.

$a+b+c$ | $4a+2b+c$ | $9a+3b+c$ | |||

1st difference | 3a+b | 5a+b | |||

2nd difference | 2a |

### The pattern-matching method

Comparing this new table to the original gives another way of figuring out the $n$th term.

6 | 17 | 34 | |||

1st difference | 11 | 17 | |||

2nd difference | 6 |

You can see that for this sequence, $2a=6$, $3a+b=11$ and $a+b+c=6$. Solving these equations gives $a=3$, $b=2$ and $c=1$. Again, the $n$th term is $3n^2+2n+1$.

To use this method you would have to memorise (or figure out) the following expressions:

- First term: $a+b+c$
- (First) first difference: $3a+b$
- (Only) Second difference: $2a$

### The simultaneous equations method

Rather than doing that, you can work out that the first term is $a+b+c$, the second term is $4a+2b+c$ and the third term is $9a+3b+c$.

This gives three simultaneous equations with three unknowns and these equations are very easy to solve.

For the example sequence 6, 17, 34… the equations are: (1) $a+b+c=6$ (2) $4a+2b+c=17$ (3) $9a+3b+c=34$

Subtracting (1) from (2) and (2) from (3) gives:

(4) $3a+b=11$ (5) $5a+b=17$

Subtracting (4) from (5) gives $2a=6$ so $a=3$.

Substituting this into equation (3) gives $9+b=11$ so $b=2$.

Substituting $a=3$ and $b=2$ into equation (1) gives $3+2+c=6$ so $c=1$.

Once again the $n$th term is found to be $3n^2+2n+1$.

The following two methods are not serious suggestions for GCSE students.

### The matrix method

In matrix form, the simultaneous equations can be written as $\begin{pmatrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} t_1 \\ t_2 \\ t_3 \end{pmatrix}$, where $t_1$, $t_2$ and $t_3$ are the first three terms of the sequence.

The values of a,b and c are given by $ \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{pmatrix}^{-1} \begin{pmatrix} t_1 \\ t_2 \\ t_3 \end{pmatrix}$

Or equivalently, $ \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} \frac {1}{2} & -1 & \frac{1}{2} \\ -\frac{5}{2} & 4 & \frac{-3}{2} \\ 3 & -3 & 1 \end{pmatrix}^{-1} \begin{pmatrix} t_1 \\ t_2 \\ t_3 \end{pmatrix}$

Checking the inverse of the matrix is left as an exercise for the reader ((+1 - Ed)).

$a=\frac{1}{2}t_1 - t_2 +\frac{1}{2}t_3$ $b=-\frac{5}{2}t_1 + 4 t_2 -\frac{3}{2}t_3$ $c= 3t_1 - 3 t_2 + t_3$

For the example sequence this is $a=\frac {1}{2}\times 6-17+\frac{1}{2}\times 34=3 - 17 + 17 = 3$ $b=-\frac{5}{2}\times 6 +4\times17-\frac{3}{2}\times 34= - 15 + 68 - 51 = 2$ $c= 3 \times 6- 3 \times 17+34= 18 - 51 + 34 = 1$

And the $n$th term is $3n^2+2n+1$.

### A special bonus method

The final method is my favourite. Imagine three quadratic functions $q_1(n)$, $q_2(n)$ and $q_3(n)$ with the following values for $n=1$, $n=2$ and $n=3$:

$q_1(1)=t_1$, $q_1 (2)=0$, and $ q_1 (3)=0$ ; $q_2(1)=0$, $q_2 (2)=t_2$, and $q_2 (3)=0$; $q_3 (1)=0$, $q_3 (2)=0$, and $ q_3 (3)=t_3$.

Add them together to get $Q(n)=q_1 (n)+q_2 (n)+q_3 (n)$.

Now $Q(1)=t_1$ , $Q(2)=t_2$ and $Q(3)=t_3$

$Q(n)$ is easy to find.

For the example question with the terms 6, 17, 34…:

$q_1(n)$ is zero at $n=2$ and $n=3$, so it’s $k(n-2)(n-3)$ for some $k$; we can find $k$ by evaluating it at $n=1$, which gives $k(-1)(-2) = 6$ so $k=3$. $q_2(n)$ turns out to be $-17 (n-1)(n-3)$, using the same idea, and $q_3(n) = 17(n-1)(n-2)$.

You can expand these, add them up and simplify down to get $3n^2 + 2n + 1$.

In general, $Q(n) = t_1 \frac{(n-2)(n-3)}{(1-2)(1-3)} + t_2 \frac{(n-1)(n-3)}{(2-1)(2-3)} + t_3 \frac{(n-1)(n-1)}{(3-1)(3-2)}$; this can be easily extended to different given terms if needed! That’s an answer that can be written down without thinking, which makes perfect sense, requires no trust and no memory. It is, of course, a bit of a mess.

Curious students may like to read about using matrices to solve simultaneous equations and Lagrange interpolation.

* Note from Colin: I welcome guest posts from people I know. The best way to get me to know you is to interact on Twitter.