 # Quadratic Sequences Questions, Worksheets and Revision

Level 6 Level 7

## What you need to know

A quadratic sequence is a sequence whose nth term formula is a quadratic. Specifically, the nth term formula for a quadratic sequence will take the form

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

where $a, b,$ and $c$ are all numbers – in GCSE maths, they will be either whole numbers or fractions. For example, you might see a sequence like

$u_n=2n^2+3$.

If you substitute values of $n$ from 1 up to 5 into this formula, you get that the first 5 terms of this sequence are

$5, 11, 21, 35, 53$.

Substituting values into a formula is easy, but the real focus of this topic is finding the nth term formula for a quadratic sequence gives the first few terms, like the list above. Before we get into an example, you need to know how to find the nth term of a linear sequence – click here (https://mathsmadeeasy.co.uk/gcse-maths-revision/sequences-linear-gcse-maths-revision-worksheets/) if you aren’t sure.

Example: Find the nth term formula of the following quadratic sequence.

$2, 9, 18, 29, 42.$

So, the nth term formula will take the form $an^2+bn+c$, where $a, b,$ and $c$ are numbers for us determine. To do this, we will first find the differences between the terms in the sequence.

We can see that unlike a linear sequence, the differences aren’t the same. However, if we then look at the differences between those differences, we see the second differences are the same.

Note: this is how you spot a quadratic sequence. If the question doesn’t specify the type of sequence, but you discover that the second differences (as we have here) are all the same, then it must be a quadratic sequence. Now, the point of doing this is that the value of $a$ (i.e. the coefficient of $n^2$ in the nth term formula) is the value of the second differences halved. So, in this case $a=1$.

We now know the nth term formula is $u_n=n^2+bn+c$, so we still need to find $b$ and $c$. To do this, we need to first write out the terms in our sequence again. Then, we need to take the part of the nth term formula that we now know, which is $n^2$ (in general it will be $an^2$), substitute the values of $n$ from 1 to 5 into this, and write the results below the original sequence. Finally, we will subtract each member of the second row from the member of the first row that sits above it and write the results in a third row. This should look like:

Here, we have called the row of differences $d_n$. Now, observe that $d_n$ was formed by subtracting the $n^2$ values from the original sequence, which has nth term $n^2+bn+c$. Moreover, if we were to actually subtract $n^2$ from $n^2+bn+c$, we’d just be left with $bn+c$. In other words, our new sequence $d_n$ is a linear sequence whose nth term will give us the missing letters of the quadratic nth term that we want. So, we find the nth term of this linear sequence as we know how to do.

Following the usual process, we see that the common difference is 4 so the nth term must be $4n+c$. To find the $c$, we observe that for $n=1$, we get that $4n=4$, whereas the first term is not 4, but 1. In order to get from 4 to 1, we must subtract 3, and so the nth term for $d_n$ is

$d_n=4n-3$.

As mentioned, this formula is precisely the last part of the quadratic nth term formula that we set out to find in the first place. In other words, $b=4$ and $c=-3$. So, recalling that we found $a$ to be 1, we get that the nth term formula for the quadratic sequence in the question is

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

Well, that was a real journey. It does seem like a long method to find the nth term of a quadratic sequence, but it is precisely that – a method. Every time you need to find the nth term formula for a quadratic sequence, the method is the same. To find $an^2+bn+c$, the process is:

– Find the difference between each term, and find the second differences (i.e. the differences between the differences);

– Check that the second differences are all the same (if they aren’t, then either something has gone wrong or it’s not a quadratic sequence), and then halve that number to get $a$;

– Write out the terms of the sequence again, then substitute the first few values of $n$ into $an^2$ and write the results below the original sequence.

– Subtract the elements of the second row from the elements above them in the first row.

– Treat these differences like a sequence and find the nth term of that linear sequence.

– Finally, combine the $an^2$ that you found in step 2 with the $bn+c$ that you found in step 5, and you’ve got your quadratic nth term formula.

### Example Questions

So, the nth term formula will take the form $an^2+bn+c$ where $a, b,$ and $c$ are numbers to be determined. Firstly, we have to find the differences between the terms in the sequences, and then find the difference between the differences. Doing so, we ge The second difference are the same, as expected, therefore $a$ is half of the second difference and so $a=1$.

Now we need to find $b$ and $c$. To do so, we need to first write out the terms in our sequence again. Then, we need to take the part of the nth term formula that we now know, which is $n^2$, substitute the values of $n$ from 1 to 5 into this, and write the results below the original sequence. Finally, we will subtract each member of the second row from the member of the first row that sits above it and write the results in a third row. This should look like: Then, $d_n$ is a linear sequence whose nth term formula is precisely $bn+c$, so finding it will give us the missing parts of our quadratic nth term formula. So, looking at the differences between the terms of $d_n$, we see The difference is 8, so the nth term must be $8n+c$. Then, for $n=1$, we get $8n=8$ which is the same as the first term of the sequence, so we must have that $c=0$ and the complete nth term formula is $d_n=8n$. Therefore, combining this with the first time in the quadratic that we found earlier, we get the nth term formula of the quadratic to be

$u_n=n^2+8n$.

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So, the nth term formula will take the form $an^2+bn+c$ where $a, b,$ and $c$ are numbers to be determined. Firstly, we have to find the differences between the terms in the sequences, and then find the difference between the differences. Doing so, we get The second difference are the same, as expected, therefore $a$ is half of the second difference and so $a=2$.

Now we need to find $b$ and $c$. To do so, we need to first write out the terms in our sequence again. Then, we need to take the part of the nth term formula that we now know, which is $2n^2$, substitute the values of $n$ from 1 to 5 into this, and write the results below the original sequence. Finally, we will subtract each member of the second row from the member of the first row that sits above it and write the results in a third row. This should look like: Then, $d_n$ is a linear sequence whose nth term formula is precisely $bn+c$, so finding it will give us the missing parts of our quadratic nth term formula. So, looking at the differences between the terms of $d_n$, we see

The difference is -5, so the nth term must be $-5n+c$. Then, for $n=1$, we get $-5n=-5$, whereas the first term is not -5, but -2. To get from -5 to -2 we have to add 3, so we must have that $c=3$, and thus the nth term for $d_n$ is

$d_n=-5n+3$ Therefore, combining this with the first time in the quadratic that we found earlier, we get the nth term formula of the quadratic to be

$u_n=2n^2-5n+3$.