The nth term of a quadratic has the form an^{2} + bn + c and to find it, we need to find the values for a, b and c

It is quite useful to notice that an nth term is made up of two distinct parts.

Quadratic Nth term = Quadratic Element + Linear Element

an^{2} + bn + c

Finding the nth term of a quadratic is quite easy, if not straightforward. Just follow these steps and you won't go far wrong.

Example: Find the nth term for the sequence 5, 15, 29, 47

**Step 1: Finding a: **

*The value of a will always be half the second difference*

The difference between the first and second terms is 10

The difference between the second and third terms is 14

The difference between the third and fourth terms is 18

We can see that the second difference is 4, so the value of a is 2

**Step 2: Set up a table**

Position | 1 | 2 | 3 | 4 |

an^{2} = 2n^{2} |
||||

bn+c | ||||

Target | 5 | 15 | 29 | 47 |

In this, the position is just the value of n (where it is in the sequence)

We have just found a so we can fill in the second line by substituting values of n into 2n^{2.}

Position | 1 | 2 | 3 | 4 |

an^{2} = 2n^{2} |
2 | 8 | 18 | 32 |

bn+c | ||||

Target | 5 | 15 | 29 | 47 |

**Step 3: Find the value of bn + c**

In our table, we now know what the quadratic elelment is giving us, so we can now find out what the linear components will be. These are just the differences between an^{2} and the target

Position | 1 | 2 | 3 | 4 |

an^{2} = 2n^{2} |
2 | 8 | 18 | 32 |

bn+c | 3 | 7 | 11 | 15 |

Target | 5 | 15 | 29 | 47 |

We can now see that these numbers in red (3, 7, 11, 15) form a linear sequence with an nth term of 4n - 1

**Step 4: Put it all together**

Earlier, we said that...

Quadratic Nth term = Quadratic Element + Linear Element

an^{2} + bn + c

** 2n ^{2} + 4n - 1 **