`6, 15, 30, 51, 78, 111.....` 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 begin finding the solution by finding the first difference: 

`x_1 = T_2 - T_1 = 15 - 6 =9`

`x_2 = T_3 - T_2 = 30 - 15 = 15`

`x_3 = T_4 - T_3 = 51 - 30 = 21`

`x_4= T_5 - T_4 = 78 - 51 = 27`

From above we can see we do not have a constant first difference, now let's find out if we have a second difference: 

`x_2- x_1 = 15 - 9 =6`

`x_3- x_2 = 21 -15 = 6`

`x_4 - x_3 = 27 - 21 = 6`

From above we have a second difference, therefore we have a constant second difference. The sequence is quadratic.

Now we know our sequence is quadratic, let's find the the model using the following equation: 

`T_n = an^2 + bn + c`

We need to find the variables a, b, c using the following equations: 

`2a =`  second difference therefore

`2a = 6`

`a = 3`

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

`3a + b = 9`

`3 (3) + b = 9` (substitute 3 for a)

`b = 9-9 =0`

Lastly: 

`a + b + c` = first term

`3 + 0 + c =6` (substitute 3 for a and 0 for b)

`c = 6 -3 = 3`

Now we have determined all three variables, lets determine the model: 

`T_n = 3n^2 + 0n +3 = 3n^2 + 3`

Now we have a model, let's double check if it is correct using term 2 and term 6:

`T_2 = 3 (2)^2 + 3 = 15`

`T_6 = 3(6)^2 +3 = 111`

SUMMARY:

Sequence: Quadratic

Model: `T_n = 3n^2 +3`