`-2, 1, 6, 13, 22, 33.....` Decide whether the sequence can be represented perfectly by a linear or a quadratic model. If so, then find the...

Firstly we need to determine whether the series is linear or quadratic. A linear sequence is a sequence of numbers in which there is a first difference between any consecutive terms is constant. However, a quadratic sequence is a sequence of numbers in which there is a second difference between any consecutive terms is constant. 

Let's determine if the above sequence has a first difference: 

`x_1 = T_2 - T_1 = 1 - (-2) = 3`

`x_2 = T_3 - T_2 = 6 - 1 = 5`

`x_3 = T_4 - T_3 = 13 - 6 = 7`

From above we observe there is no first difference, now let's determine the second difference: 

`x_2 - x_1 = 5 -3 = 2`

`x_3 - x_2 = 7 -5 = 2`

From above we observe that the second difference is constant. Therefore the sequence is perfectly quadratic.

Let's determine the quadratic model using the following formula: 

`T_n = an^2 + bn + c`

where: T_n = term value, ,n = term, variables: a,b,c

`2a` = second difference

`2a = 2`

`a = 1`

Let's find variable b: 

`3a + b =`  first difference between term 2 and term 1

`3(1) + b = 3` (substitute for a and first difference between term 2 and term1)

`b = 3-3 =0`

Lastly we are finding variable c: 

`a + b + c =` value of term 1

`1 + 0 + c = -2` (substitute for a, b and value of term 1

`c = -2 -1 = -3`

Now we have determined the variables, we can develop our model: 

`T_n = 1(n)^2 + 0(n) - 3 = n^2 - 3`

Let's double check our formula using term 4: 

`T_4 = (4)^2 - 3 = 16 -3 = 13`

SUMMARY: 

Sequence: Quadratic

Model: `T_n = n^2 - 3`