Chapter 6.7, Problem 57E

a.

To determine

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

Explanation of Solution

Given:

The given explicit rule is $a_n=a_1+(n-1)d$

Calculation:

Because this is the derivation from the explicit rule and recursive rule for the arithmetic sequence.

The explicit rule is $a_n=a_1+(n-1)d$

The recursive rule is $a_n=a_{n-1}+d$

Therefore,

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

Hence proved.

b.

To determine

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

Explanation of Solution

Given:

The explicit rule is $a_n=a_1+(n-1)d$

Calculation:

The explicit rule is $a_n=a_1+(n-1)d$

  $\begin{array}{l}{a}_n=a_1+(n-1)d\\ =a_1+[(n-1)+0]d\{\text{additive identity property}}\\ =a_1+[(n-1)-1+1]d\{\text{additive iinverse property}}\\ =a_1+[((n-1)-1)+1]d\{\text{associative property}}\\ =a_1+[(n-1)-1]d+d\{\text{distributive property}}\\ =a_{n-1}+d\text{from(a)}\end{array}$