Chapter 9.4, Problem 59E

To determine

To decide whether the sequence $-1,11,31,59,95,139$ , … can be represented perfectly be a linear or a quadratic model and then find a model.

Answer to Problem 59E

The sequence $-1,11,31,59,95,139$ , … can be represented perfectly be a quadratic model $a_n=4n^2-5$ .

Explanation of Solution

Given:

Let the sequence be $a_1=-1,{\text{ a}}_2=11,{\text{ a}}_3=31,{\text{ a}}_4=59,{\text{ a}}_5=95,{\text{ a}}_6=139,\dots$

Concept used:

• The first differences of a sequence are found by subtracting consecutive terms. The second differences are found by subtracting consecutive first differences.
• When the first differences are all the same, the sequence has a perfect linear model. That is, the sequence is arithmetic represented as $a_n=a_1+\left(n-1\right)d$ .
• When the second differences are all the same the sequence has a perfect quadratic model which is represented as $a_n=a{n}^2+bn+c$ .

Calculation:

Let’s first find out the first difference:

  $\begin{array}{l}{T}_1=a_2-a_1=11-\left(-1\right)=12\\ T_2=a_3-a_2=31-11=20\\ T_3=a_4-a_3=59-31=28\\ T_4=a_5-a_4=95-59=36\\ T_5=a_6-a_5=139-95=44\\ ⋮\end{array}$

And so on.

Since, the first differences are different, now find out second differences.

  $\begin{array}{l}{T}_2-T_1=20-12=8\\ T_3-T_2=28-20=8\\ T_4-T_3=36-28=8\\ T_5-T_4=44-36=8\\ ⋮\end{array}$

And so on.

The sequence can be represented perfectly by a quadratic model $a_n=a{n}^2+bn+c$ , since, second differences are same.

To find out the model, solve the system of equations:

For $n=1,$

  $a_1=a+b+c=-1$ … (1)

For $n=2,$

  $a_2=4a+2b+c=11$ … (2)

For $n=3,$

  $a_3=9a+3b+c=31$ … (3)

Multiply equation (1) with $-1$ and then add it to the equation (2) and (3).

  $\begin{array}{r}-a-b-c=1\\ \underset{\_}{4a+2b+c=11}\\ \left(4\right)\dots 3a+b=12\end{array}$

And,

  $\begin{array}{r}-a-b-c=1\\ \underset{\_}{9a+3b+c=31}\\ \left(5\right)\dots 8a+2b=32\end{array}$

Now multiply equation (4) with $-2$ , and then add it with equation (5).

  $\begin{array}{r}-6a-2b=-24\\ \underset{\_}{8a+2b=32}\\ \left(6\right)\dots 2a=8\end{array}$

Now solve equation (6) for $a$ and get, $a=4$ .

Now substitute $a=4$ in equation (4) and solve for $b,$ and get

  $\begin{array}{l}\Rightarrow 3\left(4\right)+b=12\\ \Rightarrow b=0\end{array}$

Finally substitute $a=4$ and $b=0$ in equation (1) and get,

  $\begin{array}{l}\Rightarrow a+b+c=-1\\ \Rightarrow 4+c=-1\\ \Rightarrow c=-5\end{array}$

Therefore, the sequence can be represented perfectly by a quadratic model $a_n=4n^2-5$ .