Quadratic Sequences Questions | Worksheets and Revision | MME

# 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. 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 to be

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

Substituting values into a formula is easy, but the real focus of this topic is finding the nth term formula for a quadratic sequence giving 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.

### The Steps to Find the Nth Term

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 steps are:

1. Find the difference between each term, and find the second differences (i.e. the differences between the differences);
2. Check that the second differences are all the same, and then halve that number to get $a$;
3. 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.
4. Subtract the elements of the second row from the elements above them in the first row.
5. Treat these differences like a sequence and find the nth term of that linear sequence.
6. 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 1: Finding the nth Term of a Quadratic Sequence

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.

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 substitute the values of $n$ from 1 to 5 into this, and write the results below the original sequence as shown.

Here, we have called the row of differences $d_n$. $d_n$ was formed by subtracting the $n^2$ values from the original sequence. $d_n$ is a linear sequence whose nth term will give us the missing letters of the quadratic nth term that we want.

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. Using this we get our final nth term to be

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

### Example 2: Finding Terms in a Quadratic Sequence

Find the first 3 terms of a sequence that has a formula $n^2+2n+5$

For this, all we need to do is find the values of $n^2+2n+5$ when n = 1, 2 and 3. To do this we substitute these values into the nth term formula

$1^2+(2\times1)+5=8$

$2^2+(2\times2)+5=13$

$3^2+(2\times3)+5=20$

So the first 3 terms of this sequence are: 8, 13, 20

### Example Questions

To generate the first 5 terms of this sequence, we will substitute $n=1, 2, 3, 4, 5$ into the formula given.

\begin{aligned}u_1 &= (1)^2+6(1)-10=-3 \\ u_2&=(2)^2+6(2)-10=6 \\ u_3&=(3)^2+6(3)-10=17 \\ u_4&=(4)^2+6(4)-10=30 \\ u_5&=(5)^2+6(5)-10=45\end{aligned}

Hence the first five terms of the sequence are,

$-3, 6, 17, 30, 45$

a) To generate the first 4 terms of this sequence, we will substitute $n=1, 2, 3, 4$ into the formula given.

\begin{aligned}u_1 &=(1)^2-2 =-1 \\ u_2 &=(2)^2-2 =4 \\ u_3 & =(3)^2-2=7 \\ u_4 &=(4)^2-2=14 \end{aligned}

b) Every term in this sequence is generated when an integer value of n is substituted into $n^2-2$

Hence if we set 287 to equal $n^2-2$, we can determine its position in the sequence,

$n^2-2=287$

making n the subject,

$n=\sqrt{287+2}=17$

Hence 287 is the $17^{th}$ term in the sequence.

Every term in this sequence is generated when an integer value of n is substituted into $(n-1)^2$

Thus if we set 49 to equal $(n-1)^2$, we can determine its position in the sequence,

\begin{aligned}(n-1)^2&=49 \\ n-1 &=\pm\sqrt{49} \\ n&=1 +7 = 8\end{aligned}

Hence 49 is the $8^{th}$ term in the sequence, as n can only be positive integers.

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 find,

The second difference is the same, as expected, therefore $a$ is half of the second difference and so $a=1$

Now we need to find $b$ and $c$ by comparing the values generated by a sequence of $n^2$, to the original sequence.

\begin{aligned} u_n & = 9 , 20, 33, 48, 65 \\ n^2 & = 1, 4, 9, 16, 25 \\ d_n & = 8, 16, 24, 32, 40\end{aligned}

The difference, $d_n$, is a linear sequence whose nth term formula is precisely $bn+c$, so

The difference is 8, so the nth term must be $8n+c$ where $c=0$

Therefore, we get the nth term formula of the quadratic to be,

$u_n=n^2+8n$

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 find,

The second difference is the same, as expected, therefore $a$ is half of the second difference and so $a=2$

Now we need to find $b$ and $c$ by comparing the values generated by a sequence of $2n^2$, to the original sequence.

\begin{aligned} u_n & = 0, 1, 6, 15, 28 \\ 2n^2 & = 2, 8, 18, 32, 50 \\ d_n & = -2, -7, -12, -17, -22\end{aligned}

The difference, $d_n$, is a linear sequence whose nth term formula is precisely $bn+c$, so

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

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

Level 6-7

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