# Quadratic, Cubic and Harder Sequences

## Quadratic, Cubic and Harder Sequences Revision

**Quadratic, Cubic and Harder Sequences**

Following on from **linear sequences**, which have a **common difference** between each term, the sequences found on this page are a bit more complicated, as you have recognise the **difference between the differences**.

For **quadratic** and **cubic **sequences you will need to work out the second and third difference respectively.

**Simple Quadratic and Cubic Sequences**

For certain simpler **quadratic **and **cubic **sequences we can find the nth term by comparing to the list of **square **and **cubic **numbers.

**Example: **

Find the \textcolor{#10a6f3}{n}**th** term for the quadratic sequence with the following first 5 terms:

\textcolor{#10a6f3}{3,6,11,18,27}

We can see that the differences between each term are **not equal**.

Comparing with the sequence \textcolor{#10a6f3}{n^2}, each term in our sequence is \textcolor{#10a6f3}{2} greater than the corresponding terms in the sequence n^2.

Therefore, we can conclude that the nth term of this sequence is \textcolor{#10a6f3}{n^2+2}.

**Quadratic Sequences**

For **harder** **quadratic** sequences you will be required to work out the **second difference** (or differences between the differences) to work out the **coefficient **of the n^2 term in the nth term.

You can then subtract the sequence \textcolor{#00bfa8}{a}n^2 (where \textcolor{#00bfa8}{a} is to be found) term by term away from your original sequence, leaving you with a **linear sequence** to work with.

**Example:**

Find the nth term of the sequence with the following first 5 terms

6,8,14,24,38

Firstly we need to find the **differences **until they are **constant**.

In this case we can see that the **first** **differences** are +2,+6,+10,+14 and then the **second difference** is \textcolor{#10a6f3}{+4}.

The **coefficient** of the n^2 term of a quadratic sequence is half of the second difference, which in this case is 4\div2=\textcolor{#00bfa8}{2}.

We then subtract the sequence with nth term 2n^2 away from our original sequence, u_n.

This leaves us with a **linear sequence**, which has a term to term difference of -4 and therefore an nth term of \textcolor{#10a6f3}{-4n+8}.

Thus the nth term to the quadratic sequence is \textcolor{#10a6f3}{2n^2-4n+8}.

**Cubic Sequences**

To find the nth term of a **cubic sequence** we will need to find the **first, second and third differences**.

Similar to a **quadratic sequence** you then subtract an^3 from your original sequence, leaving you with a **quadratic sequence**, where you then follow the same steps as above to find the nth term.

**Example:**

Find the nth term of the sequence with the following 5 terms

1,28,101,238,457

We need to first find the **differences** until they are constant.

In this case we can see that the **first differences** are \textcolor{#10a6f3}{+27,+73,+137,+219}, the **second differences** are \textcolor{#10a6f3}{+46,+64,+82} and then the **third difference** is \textcolor{#10a6f3}{+18}.

The **coefficient** of the n^3 term is \textcolor{#00bfa8}{\dfrac{k}{6}}, where \textcolor{#00bfa8}{k} is the **constant third difference**.

Subtracting 3n^3 away from the sequence:

We are now left with a **quadratic sequence** with a **second constant difference** of +10.

So the **coefficient** of n^2 is 10\div 2=\textcolor{#00bfa8}{5}.

Subtracting 5n^2 away from u_n-3n^3.

This now leaves with us a **linear sequence** with an nth term of \textcolor{#10a6f3}{-9n+2}.

Hence the nth term of this **cubic sequence** is \textcolor{#10a6f3}{3n^3+5n^2-9n+2}.

**Exponential Sequences**

**Exponential sequences** increase by a common **ratio **from term to term, with the nth term being defined as

where \textcolor{#d11149}{a} is the **first term** and \textcolor{#00d865}{r} is the value that you **multiply **by each time.

**Example:**

An exponential sequence x_n has the first 5 terms

5,15,45,135,405

Find the nth term of the sequence.

To find \textcolor{#00d865}{r}, divide the next term by the previous term.

\textcolor{#00d865}{r}=\dfrac{x_{n+1}}{x_n}=\dfrac{15}{5}=\textcolor{#00d865}{3}

Therefore the nth term is \textcolor{#d11149}{5}\times \textcolor{#00d865}{3}^{(n-1)}, as \textcolor{#d11149}{a=5}.

## Quadratic, Cubic and Harder Sequences Example Questions

**Question 1:** A quadratic sequence has the following first 5 terms

-2,1,6,13,22

Find the nth term of the sequence and hence the 25th term.

**[3 marks]**

Comparing with the sequence n^2, we can see that each term is 3 less.

So the nth term is n^2-3.

Substituting in n=25 to find the 25th term:

25^2-3=622**Question 2:** Find the nth term of the quadratic sequence with the following first 5 terms

9,25,47,75,109

**[4 marks]**

The first differences between terms are +16,+22,+28,+34

The second difference between terms is +6

Therefore, the coefficient of the n^2 term is 6\div 2=3

Subtracting 3n^2 away from the sequence term by term gives us the following linear sequence

6,13, 20,27,34

The nth term of this sequence is 7n-1

Thus, the nth term of the quadratic sequence is 3n^2+7n-1

**Question 3:** Find the next term of the cubic sequence with the following first 5 terms

4,9,28,73,156

**[2 marks]**

To work out the next term in this sequence we need to find the first, second and third differences.

First differences: +5,+19,+45,+83

Second differences: +14,+26,+38

Third difference: +12

Working backwards we know that the next second difference will be +50 and therefore the next first difference will be +133.

Therefore, the next term in this sequence is 156+133=289

**Question 4:** Find the nth term of the cubic sequence with the following first 5 terms

1,6,17,40,81

**[5 marks]**

First differences: +5,+11,+23,+41

Second differences: +6, +12,+18

Third difference: +6

Coefficient of the n^3 term =6\div6=1

Subtracting n^3 away from the sequence leaves us with the following sequence

0,-2,-10,-24,-44

The first differences of this quadratic sequence are -2,-8,-14,-20 and the second difference is -6

Therefore, the n^2 coefficient is -6\div 2=-3

Subtracting -3n^2 away from the quadratic sequence gives us the linear sequence

3, 10,17,24,31

The nth term of this linear sequence is 7n-4

Putting everything together gives us the nth term of our cubic sequence, which is n^3-3n^2+7n-4

**Question 5:** An exponential sequence has the following first 4 terms

3,12,48,192

a) Find the nth term of this sequence

**[2 marks]**

b) Hence or otherwise, find the 10th term of this sequence.

**[1 mark]**

a) Work out the common ratio, r from term to term.

r=\dfrac{12}{3}=4

Therefore the nth term is 3\times 4^{(n-1)}

b) Substituting n=10 gives

3\times 4^{(10-1)}=786432