Chapter 7.2, Problem 66PS

Solution

i.

To identify: The table of values showing d(1), d(2), d(3), and d(4)

Making a Table Let d(n) represent the distance the object falls in the nth second (Hint: The distance d(1) that the object falls in the first second is D(1) - D(0).)

The distance D (in feet) that an object falls in t seconds can be modeled by D(t)= 16

The required rule is $a_n=8n-4$ .

Given information:

  $\begin{array}{l}d\left(n\right)=D\left(n\right)-D\left(n-1\right)\\ D\left(t\right)=16t^2\end{array}$

Explanation:

It is given that:

  $\begin{array}{l}d\left(n\right)=D\left(n\right)-D\left(n-1\right)\\ D\left(t\right)=16t^2\end{array}$

Put the value of $1,2,3,4$ into

  $\begin{array}{l}d\left(n\right)=D\left(n\right)-D\left(n-1\right)\\ D\left(t\right)=16t^2\end{array}$

Put different values of n ,

  $\begin{array}{c}d\left(1\right)=D\left(1\right)-D\left(0\right)\\ D\left(1\right)-D\left(0\right)=16\cdot 1^2-16\cdot 0^2\\ 16\cdot 1^2-16\cdot 0^2=16-0\\ 16-0=16\\ D\left(2\right)-D\left(1\right)=16\cdot 2^2-16\cdot 1^2\\ 16\cdot 4-16\cdot 1=48\\ D\left(3\right)-D\left(2\right)=16\cdot 3^2-16\cdot 2^2\\ 16\cdot 9-16\cdot 4=80\\ D\left(4\right)-D\left(3\right)=16\cdot 4^2-16\cdot 3^2\\ 16\cdot 16-16\cdot 9=112\end{array}$

The table of values.

  $\begin{array}{ccccc}n& 1& 2& 3& 4\\ d_n& 16& 48& 80& 112\end{array}$

ii

To identify: The rule for the sequence of distances given by $\left(d_n\right)$ .

The rule for the sequence of distance is $\left(d_n\right)=32n-16$

Given information:

  $\begin{array}{l}d\left(n\right)=D\left(n\right)-D\left(n-1\right)\\ D\left(t\right)=16t^2\end{array}$

Explanation:

It is given term

  $\begin{array}{l}d\left(n\right)=D\left(n\right)-D\left(n-1\right)\\ D\left(t\right)=16t^2\end{array}$

Using the from part (1)

The general term is $d\left(n\right)=D\left(n\right)-D\left(n-1\right)$ .

Product platform rewrite

  $\begin{array}{l}\left(d_n\right)=16{\left(n\right)}^2-16{\left(n-1\right)}^2\\ \left(d_n\right)=16n^2-16\left(n^2-2n+1\right)\\ \left(d_n\right)=16n^2-16n^2+32-16\\ \left(d_n\right)=32n-16\end{array}$

Therefore, the rule for the sequence of distance is $\left(d_n\right)=32n-16$

iii

To identify: Drawing a graph the sequence from part (2 )

  

Given information:

  $\begin{array}{l}d\left(n\right)=D\left(n\right)-D\left(n-1\right)\\ D\left(t\right)=16t^2\end{array}$

Explanation:

It is given term

  $\begin{array}{l}d\left(n\right)=D\left(n\right)-D\left(n-1\right)\\ D\left(t\right)=16t^2\end{array}$

Using the equation from part (2) $\left(d_n\right)=32n-16$ several points to plot by entering several n values and solving for $\left(d_n\right)$

Put the value of $n$ into $\left(d_n\right)=32n-16$

  $\begin{array}{c}d\left(n\right)=-16+32n\\ d\left(1\right)=-16+32\left(1\right)\\ -16+32\left(1\right)=16\\ d\left(2\right)=-16+32\left(2\right)\\ -16+32\left(2\right)=48\\ d\left(3\right)=-16+32\left(3\right)\\ -16+32\left(3\right)=80\\ d\left(4\right)=-16+32\left(4\right)\\ -16+32\left(4\right)=112\\ \\ \begin{array}{cc}n& d_n\\ 1& 16\\ 2& 48\\ 3& 80\\ 4& 112\end{array}\end{array}$

Draw the graph.