Chapter 6.6, Problem 57E

(a)

To determine

To explain $a_{n-1}=a_1+\left[\left(n-1\right)-1\right]d$

Explanation of Solution

Given:

The explicit rule $a_n=a_1+\left(n-1\right)d$ defines an arithmetic sequence

Concept Used:

Recursive equation for an arithmetic sequence is

  $a_n=a_{n-1}+d\text{ (where d is common difference)}$

Calculation:

Recursive rule is

  $a_n=a_{n-1}+d\text{ (where d is common difference)}$

Find value of $a_{n-1}$

  $a_{n-1}=a_n-d\mathrm{.....}\left(1\right)$

Given explicit rule $a_n=a_1+\left(n-1\right)d$

Put value of $a_n$ in equation (1)

  $\begin{array}{c}{a}_{n-1}=a_1+\left(n-1\right)d-d\\ =a_1+\left[\left(n-1\right)-1\right]d\text{ (taking }d\text{ common)}\end{array}$

Conclusion:

Hence,

  $a_{n-1}=a_1+\left[\left(n-1\right)-1\right]d$ explained

(b)

To determine

To justify each step showing that a recursive equation for the sequence is $a_n=a_{n-1}+d$

Explanation of Solution

Given:

Steps showing that a recursive equation for the sequence is $a_n=a_{n-1}+d$ are

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)d\\ =a_1+\left[\left(n-1\right)+0\right]d\\ =a_1+\left[\left(n-1\right)-1+1\right]d\\ =a_1+\left[\left(\left(n-1\right)-1\right)+1\right]d\\ =a_1+\left(\left(n-1\right)-1\right)d+d\\ =a_{n-1}+d\end{array}$

Concept Used:

Use $a_{n-1}=a_1+\left[\left(n-1\right)-1\right]d$ to justify

Calculation:

Below is the justification for each step

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)d\\ \left(\text{addition of zero to any term does not change it }x=x+0=x\right)\\ a_n=a_1+\left[\left(n-1\right)+0\right]d\\ \left(\text{zero can be written as sum of a number and its additive inverse example here }0=-1+1\right)\\ a_n=a_1+\left[\left(n-1\right)-1+1\right]d\\ \text{(use associative property }\left(\left(x+y\right)+\left(z+h\right)\right)\text{=}\left(\left(x+y\right)\text{+z}\right)\text{+h) }\\ =a_1+\left[\left(\left(n-1\right)-1\right)+1\right]d\\ \left(\text{use distributive property to open the bracket }x\times \left(y+z\right)=\left(x\times y\right)\text{+}\left(x\times \text{z}\right)\right)\\ =a_1+\left(\left(n-1\right)-1\right)d+d\\ \left(\text{subsitute }{a}_{n-1}=a_1+\left[\left(n-1\right)-1\right]d\right)\\ =a_{n-1}+d\end{array}$