Precalculus, Chapter 9, 9.4, Section 9.4, Problem 59

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