Chapter 9.2, Problem 47PPSE

Solution

To write: An explicit and a recursive formula for the sequence $1,1\frac{1}{3},1\frac{2}{3},2,\dots$ .

The explicit formula is $a_n=\frac{n+2}{3}$ and the recursive formula is $a_n=a_{n-1}+\frac{1}{3},a_1=1$ .

Given information:

The sequence $1,1\frac{1}{3},1\frac{2}{3},2,\dots$ .

Concept used:

Difference between two consecutive terms of an arithmetic sequence is common and the difference is called common difference.

The nth term of an arithmetic sequence is given by $a_n=a+\left(n-1\right)d$ , where $a$ is the first term and $d$ is the common difference.

The $n\text{th}$ term of an arithmetic sequence can be found by adding the common difference to $\left(n-1\right)\text{th}$ term. Therefore, the recursive formula can be given as $a_n=a_{n-1}+d$ .

Calculation:

The first term of the sequence is $a=1$ and the common difference is $d=\frac{1}{3}$ .

Substitute these values in the formula, $a_n=a+\left(n-1\right)d$ , to find the explicit formula.

  $\begin{array}{c}{a}_n=1+\left(n-1\right)\frac{1}{3}\\ =1+\frac{n}{3}-\frac{1}{3}\\ =\frac{n}{3}+\frac{2}{3}\\ =\frac{n+2}{3}\end{array}$

So, the explicit formula is $a_n=\frac{n+2}{3}$ .

Since $d=\frac{1}{3}$ , the recursive formula for the sequence is $a_n=a_{n-1}+\frac{1}{3},a_1=1$ .

Conclusion:

The explicit formula is $a_n=\frac{n+2}{3}$ and the recursive formula is $a_n=a_{n-1}+\frac{1}{3},a_1=1$ .