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# Quadratic Sequences

11

Sequences

Algebra

###### Quadratic sequences take the form:

###### If we substitute the first term (1) into the general form, we getâ€¦

###### Which simplifies toâ€¦

###### This is shown in the table below immediately beneath the termâ€™s value.

###### Here we have a sequence of numbers. I have included the differences between the numbers in the sequence.

###### Notice that the 2nd difference is constant across the sequence- a sure sign that you have a quadratic sequence.

###### Finding the next number in the sequence is easy - add the 1st and second differences together from n=4 to n=5 (58 + 8 = 66) then add this to the value for n=5 (207 + 66 = 273). The value for n=6 is 273.

###### Let's find out the quadratic equation that gave this sequence of numbers...

###### From the derived equations in the first table, we can calculate the values for a, b and c.

###### Sequence = a + b + c = **23**

###### 1st difference = 3a + b = **34**

###### 2nd difference = 2a = **8**

###### Of course, you can memorise the formulas, but they can still be worked out if you forget.

###### For a:

###### 2a = 8

###### So a = 4

###### Substitute this value for a in 3a + b = 34

###### 3(4) + b = 34

###### So b = 22

###### Finallyâ€¦

###### a + b + c = 23

###### 4 + 22 + c = 23

###### c = -3

###### We can put these values into the quadratic equation and, voila, get...

###### Make sure you check it with a value for n, for example n = 2... we should get 57, the second number in the sequence.

###### What if we need to find a value in the sequence, perhaps the number for when n = 20? Just substitute into the equation...

###### When n = 20, the number in the sequence is 2037.

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