Precalculus, Chapter 9, 9.4, Section 9.4, Problem 57

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