Precalculus, Chapter 9, 9.4, Section 9.4, Problem 60

The given sequence is:
-1, 8 , 23, 44, 71, 104
To determine if it is a linear sequence, take the difference between consecutive terms.
-1, 8 , 23, 44, 71, 104
vvv vvv vvv vvv vvv
9 1521 27 33
Notice that the result are not the same. So the sequence is not linear.
Then, take the second difference of the consecutive terms to determine if it is quadratic.
9, 15, 21, 27, 33
vvv vvv vvv vvv
6 6 6 6
Since the second difference of the consecutive terms are the same, the sequence is quadratic.
To determine the model of the quadratic expression apply the formula:
T_n = an^2 + bn + c
where Tn is the nth term of the sequence.
To solve for the values of a, b and c, consider the first few three terms of the sequence.
Plug-in T1=-1 and n=1.
-1=a(1)^2+b(1)+c
-1=a+b+c (Let this be EQ1.)
Also, plug-in T2=8 and n=2.
8=a(2)^2 + b(2) + c
8=4a+2b+c (Let this be EQ2.)
And, plug-in T3=23 and n=3.
23=a(3)^2+b(3)+c
23=9a+3b+c (Let this be EQ3.)
Then, apply elimination method of system of equations. Let's eliminate variable c.
To eliminate c, subtract EQ2 from EQ1.
EQ1: -1=a+b+c
EQ2: -(8=4a+2b+c)
-----------------
-9 = -3a -b
And simplify the resulting equation.
-9=-3a-b
9=3a+b (Let this be EQ4.)
Eliminate c again. So subtract EQ3 from EQ1.
EQ1: -1=a+b+c
EQ2: -(23=9a+3b+c)
-----------------
-24=-8a-2b
And, simplify the resulting equation.
-24=-8a-2b
12=4a+b (Let this be EQ5.)
Then, eliminate another variable. Let it be c.
To eliminate c, subtract EQ4 from EQ5.
EQ5: 12=4a+b
EQ4: -(9=3a+b)
--------------
3=a
Now that the value of a is known, plug-in it to either EQ4 or EQ5. Let's use EQ4.
9=3a+b
9=3(3)+b
0=b
Then, plug-in the value of a and b to either EQ1, EQ2 or EQ3. Let's use EQ1.
-1=a+b+c
-1=3+0+c
-4=c
Now that values of a, b and c, plug-in them to the formula of nth term of quadratic expression.
T_n=an^2+bn+c
T_n=3n^2+(0)n+(-4)
T_n=3n^2-4
Therefore, the given -1, 8 , 23, 44, 71, 104 is a quadratic sequence. The model for its nth term is T_n=3n^2-4 .