- Unlike in a linear sequence, in a quadratic sequence the differences between the terms (the
**first differences**) are not constant - However, the differences between the differences (the
**second differences**) are constant - Another way to think about this is that in a quadratic sequence, the
**sequence of first differences is a linear sequence**eg Sequence: 2, 3, 6, 11, 18, …

1st Differences: 1 3 5 7 (a Linear Sequence)2nd Differences: 2 2 2 (Constant)

- If the
**second differences there are constant**, we know that the example is a**quadratic sequence**

- You should be able to recognise and continue a quadratic sequence
- You should also be able to find a formula for the
*n*^{th}term of a quadratic sequence in terms of*n* - This formula will be in the form:

(The process for finding*n*^{th}term =*an*^{2}+*bn*+*c**a*,*b*, and*c*is given below)

- Work out the sequences of first and second differences
Note: check that the first differences are not constant and the second differences are constant, to make sure you have a quadratic sequence!

- e.g. sequence: 1, 10, 23, 40, 61
- first difference: 9, 13, 17, 21, ...
- second differences: 4, 4, 4, ...

- e.g. sequence: 1, 10, 23, 40, 61
*a*= [the second difference] ÷ 2- e.g.
*a*= 4 ÷ 2 = 2

- e.g.
- Write out the first three or four terms of
*a**n*^{2}with the first three or four terms of the given sequence underneath.

Work out the difference between each term of*an*^{2}and the corresponding term of the given sequence.- e.g.
*an*^{2}= 2*n*^{2}= 2, 8, 18, 32, ...

sequence = 1, 10, 23, 40, ...

difference = -1, 2, 5, 8, ...

- e.g.
- Work out the linear
*n*th term of these differences. This is*bn*+*c*.- e.g.
*bn*+*c*= 3*n*− 4

- e.g.
- Add this linear
*n*th term to*an*^{2}. Now you have*an*+^{2}*bn*+*c.*- e.g.
*an*+^{2}*bn*+*c*= 2*n*^{2}+ 3*n*− 4

- e.g.

- Before doing the very formal process to find the
*n*^{th }term, try comparing the sequence to the square numbers 1, 4, 9, 16, 25, … and see if you can spot the formula - For example:
- Sequence 4, 7, 12, 19, 28, …
- Square Numbers 1 4 9 16 25
- We can see that each term of the sequence is 3 more than the equivalent square number so the formula is
*n*^{th}term =*n*^{2 }+ 3 - This could save you a lot of time!

For the sequence 5, 7, 11, 17, 25, ....

(a)

Find a formula for the *n*^{th }term.

Start by finding the first and second differences

Sequence: 5, 7, 11, 17, 25

First differences: 2, 4, 6, 8, ...

Second difference: 2, 2, 2, ...

Hence

Now write down *a**n*^{2} (just *n*^{2} in this case as *a* = 1) with the sequence underneath, and on the next line write the difference between *a**n*^{2} and the sequence

sequence: 5, 7, 11, 17, ...

difference: 4, 3, 2, 1, ...

Work out the

Add *a**n*^{2} and *bn* + *c *together to give you the *n*^{th} term of the sequence

(b)

Hence find the 20^{th} term of the sequence.

Substitute*n* = 20 into *n*^{2} − *n* + 5

Substitute

(20)^{2} − 20 + 5 = 400 − 15

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