Chapter 9.4, Problem 60E

To determine

To decide whether the sequence $-2,13,38,\text{ 73},\text{ 118},173$ , … can be represented perfectly be a linear or a quadratic model and then find a model.

Answer to Problem 60E

The sequence $-2,13,38,\text{ 73},\text{ 118},173$ , … can be represented perfectly be a quadratic model $a_n=5n^2-7$ .

Explanation of Solution

Given:

Let the sequence be $a_1=-2,{\text{ a}}_2=13,{\text{ a}}_3=38,{\text{ a}}_4=73,{\text{ a}}_5=118,{\text{ a}}_6=173,\dots$

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:

  $\begin{array}{l}{T}_1=a_2-a_1=13-\left(-2\right)=15\\ T_2=a_3-a_2=38-13=25\\ T_3=a_4-a_3=73-38=35\\ T_4=a_5-a_4=118-73=45\\ T_5=a_6-a_5=173-118=55\\ ⋮\end{array}$

And so on.

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

  $\begin{array}{l}{T}_2-T_1=25-15=10\\ T_3-T_2=35-25=10\\ T_4-T_3=45-35=10\\ T_5-T_4=55-45=10\\ ⋮\end{array}$

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 $n=1,$

  $a_1=a+b+c=-2$ … (1)

For $n=2,$

  $a_2=4a+2b+c=13$ … (2)

For $n=3,$

  $a_3=9a+3b+c=38$ … (3)

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

  $\begin{array}{r}-a-b-c=2\\ \underset{\_}{4a+2b+c=13}\\ \left(4\right)\dots 3a+b=15\end{array}$

And,

  $\begin{array}{r}-a-b-c=2\\ \underset{\_}{9a+3b+c=38}\\ \left(5\right)\dots 8a+2b=40\end{array}$

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

  $\begin{array}{r}-6a-2b=-30\\ \underset{\_}{8a+2b=40}\\ \left(6\right)\dots 2a=10\end{array}$

Now solve equation (6) for $a$ and get, $a=5$ .

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

  $\begin{array}{l}\Rightarrow 3\left(5\right)+b=15\\ \Rightarrow b=0\end{array}$

Finally substitute $a=5$ and $b=0$ in equation (1) and get,

  $\begin{array}{l}\Rightarrow a+b+c=-2\\ \Rightarrow 5+c=-2\\ \Rightarrow c=-7\end{array}$

Therefore, the sequence can be represented perfectly by a quadratic model $a_n=5n^2-7$ .