# Ask Uncle Colin: Quadratic sequences

Dear Uncle Colin,

I had to find the $n$th term of a quadratic sequence (1, 6, 17, 34, 57). I remember my teacher saying something about a table, but I couldn’t figure it out. Can you help?

Struggles Expressing Quadratics Using Educator’s Notation - Concrete Explanation?

Hi, SEQUENCE, and thank you for your message!

(Before beginning, I should point you at Mark’s article on this topic from some time ago.)

I have used table-based methods in the past, but - just like you say - they don’t seem to stick in the memory. Instead, I much prefer an algebra-based solution. It’s a little harder conceptually, but much less to remember.

### An algebraic approach

The first thing you need is a template solution. You know that the sequence is quadratic, so the $n$th term will be of the form $an^2 + bn + c$, for some values of $a$, $b$ and $c$ that you need to find.

So, the first term is $a + b + c$; the second term is $4a + 2b+ c$; the third term is $9a + 3b + c$ - just by replacing the $n$ with 1, 2 and then 3.

That gives us three equations:

- $a+b+c = 1$
- $4a + 2b + c = 6$
- $9a + 3b + c = 17$

Subtracting the first two gives $3a + b = 5$; subtracting the second two gives $5a + b = 11$.

Subtracting those two gives $2a = 6$, so $a=3$ - without very much effort at all!

We also know that $3a+b=5$, so if $a=3$, $b= -4$.

Lastly, $a+b+c=1$, so $3 - 4 + c = 1$, giving $c=2$.

Replacing those values in the template, our $n$th term is $3n^2 - 4n + 2$.

### A slightly quicker way

If you’re prepared to be a bit sneaky, you can cut your work significantly by thinking about the 0th term.

You’d presumably be perfectly happy finding the seventh term of the sequence - the consecutive differences are 5, 11, 17, and 23 - so the next one will be 29 more than 57, which is 86.

The 0th term uses the same technique, but going the other way! The difference from the 0th term to the 1st term is -1 (six less than five), so the 0th term is 2.

Looking at the template again, the 0th term would be… just $c$! This gives us $c=2$ without any algebra at all.

We do still need to do a bit of algebra to find $a$ and $b$, though: we have $a + b + 2 = 1$ and $4a + 2b + 2 = 6$ as before (but with $c$ replaced).

These tidy up to $a+b=-1$ and $4a + 2b = 4$ - or better, $2a+b=2$ when you halve it all.

Subtracting those give $a=3$; substituting into $a+b=-1$ gives $b=-4$, as before.

Hope that helps!

- Uncle Colin