Here is a summary of all of this (without having to use the nightmare equation!)

Method 1 for More Simple Quadratics

Tonight so far we looked at quadratics of the form an2 + b;

e.g.

First sequence / 4 / 13 / 28 / 49 / 76
Difference / 9 / 15 / 21 / 27
2nd difference / 6 / 6 / 6

We know its quadratic and that it has something to do with 3n2 so we write out the 3n2 ‘times’ table as:

n = 1 / 3n2 / 3
n = 2 / 3n2 / 12
n = 3 / 3n2 / 27
n = 4 / 3n2 / 48
n = 5 / 3n2 / 75

We insert the original sequence then we put the 3n2 column as the fourth column.

Original / 3n2
n = 1 / 3n2 / 4 / 3
n = 2 / 3n2 / 13 / 12
n = 3 / 3n2 / 28 / 27
n = 4 / 3n2 / 49 / 48
n = 5 / 3n2 / 76 / 75

Doing the subtraction of the 3n2 times table from the original sequence give the difference in the 5th column.

Original / 3n2 / Difference
n = 1 / 3n2 / 4 / 3 / 1
n = 2 / 3n2 / 13 / 12 / 1
n = 3 / 3n2 / 28 / 27 / 1
n = 4 / 3n2 / 49 / 48 / 1
n = 5 / 3n2 / 76 / 75 / 1

All of the numbers in our sequence are 1 greater than the numbers in the 3n2 times table, so our formula is

3n2 + 1

Moving on……to find quadratics of the form an2 + bn + c

What do we do when something is more complex than in method 1? e.g.

Method 2 for more Complex Quadratic Sequences

First sequence / 2 / 7 / 16 / 29 / 46
Difference / 5 / 9 / 13 / 17
2nd difference / 4 / 4 / 4

We knowit’s quadratic and that it has something to do with 2n2 so we write out the 2n2 ‘times’ table as:

n = 1 / 2n2 / 2
n = 2 / 2n2 / 8
n = 3 / 2n2 / 18
n = 4 / 2n2 / 32
n = 5 / 2n2 / 50

We insert the original sequence then we put the 2n2 column as the fourth column.

Original / 2n2
n = 1 / 2n2 / 2 / 2
n = 2 / 2n2 / 7 / 8
n = 3 / 2n2 / 16 / 18
n = 4 / 2n2 / 29 / 32
n = 5 / 2n2 / 46 / 50

Doing the subtraction of the 2n2 times table from the original sequence give the difference in the 5th column.

Original / 2n2 / Difference
n = 1 / 2n2 / 2 / 2 / 0
n = 2 / 2n2 / 7 / 8 / -1
n = 3 / 2n2 / 16 / 18 / -2
n = 4 / 2n2 / 29 / 32 / -3
n = 5 / 2n2 / 46 / 50 / -4

The numbers in this sequence increase in difference by 1 each time. However, we then take the 2nd difference of these differences which looks like:

Original / 2n2 / Difference / 2nd Differences
n = 1 / 2n2 / 2 / 2 / 0
-1
n = 2 / 2n2 / 7 / 8 / -1
-1
n = 3 / 2n2 / 16 / 18 / -2
-1
n = 4 / 2n2 / 29 / 32 / -3
-1
n = 5 / 2n2 / 46 / 50 / -4

This time, we do not divide these 2nd differences by 2 to get the term ‘bn’, we just leave it as it is so that 1 means n, 2 means 2n, 5 means 5n and -1 means -1n.

So we now have the formula 2n2 – n, so we can compare it with our original formula yet again by drawing another table which uses this new formula instead of 2n2 and it gives us:

Original / 2n2 - n / Difference
n = 1 / 2n2 / 2 / 1 / 1
n = 2 / 2n2 / 7 / 6 / 1
n = 3 / 2n2 / 16 / 15 / 1
n = 4 / 2n2 / 29 / 28 / 1
n = 5 / 2n2 / 46 / 45 / 1

Subtracting all the 2n2 - n terms from the original terms now gives a common difference of 1.

Therefore the final formula is 2n2– n + 1

You might notice a pattern emerging between Methods 1 and 2 now. The object of this methodology, however complex the quadratic, is to reduce the difference column to one which is a constant difference between the new sequences and the originals.

Also, we did not have to use the a + (n-1)d + (½n2 - 1½n + 1) C equation.

Now turn over to see Method 3 which is explains an alternative method that I was bumbling about earlier, and still means that you do not have to use the nightmare equation above.

Method 3 for more Complex Quadratic Sequences – but its an easier alternative

Using the example from Method 2, we have:

First sequence / 2 / 7 / 16 / 29 / 46
Difference / 5 / 9 / 13 / 17
2nd difference / 4 / 4 / 4

As before, we know it’s quadratic and that it has something to do with 2n2 so we write out the 2n2 ‘times’ table as:

n = 1 / 2n2 / 2
n = 2 / 2n2 / 8
n = 3 / 2n2 / 18
n = 4 / 2n2 / 32
n = 5 / 2n2 / 50

We then do the usual thing and insert the original sequence then we put the 2n2 column as the fourth column.

Original / 2n2
n = 1 / 2n2 / 2 / 2
n = 2 / 2n2 / 7 / 8
n = 3 / 2n2 / 16 / 18
n = 4 / 2n2 / 29 / 32
n = 5 / 2n2 / 46 / 50

Doing the subtraction of the 2n2 times table from the original sequence give the difference in the 5th column.

Original / 2n2 / Difference
(New sequence)
n = 1 / 2n2 / 2 / 2 / 0
n = 2 / 2n2 / 7 / 8 / -1
n = 3 / 2n2 / 16 / 18 / -2
n = 4 / 2n2 / 29 / 32 / -3
n = 5 / 2n2 / 46 / 50 / -4

However, this is where it differs from Method 2. We now put have a new sequence which we can apply the equationdn + (a – d) to, where a = 0, and d = -1.

Substituting in for a and d gives -1n + (0 - -1) = -n + 1

.. and add this onto 2n2 gives 2n2 – n + 1

PTO for some more fun!

Try each method yourself on:-

If you like Method 3 you can try it on, or try the other methods on this until you feel comfortable with it.

First sequence / 4 / 10 / 22 / 40 / 64
Difference / 6 / 12 / 18 / 24
2nd difference / 6 / 6 / 6

Maybe try another:

First sequence / 6 / 17 / 34 / 57 / 86
Difference / 11 / 17 / 23 / 29
2nd difference / 6 / 6 / 6