Chapter 12.1, Problem 16E

To determine

Calculate the first five term of the given sequence.

Answer to Problem 16E

The first five term of the sequence are $1,\frac{1}{2},\frac{2}{3},\frac{3}{5},\frac{5}{8}$ .

Explanation of Solution

Given:

It is given in the question that the sequence is $a_n=\frac{1}{1+a_{n-1}}and{a}_1=1$ .

Concept Used:

In this, use the concept of arithmetic progression in which put the number and calculate the problem.

Calculation:

Here, the sequence is $a_n=\frac{1}{1+a_{n-1}}and{a}_1=1$ .

First, find $a_2$ ,

  $\begin{array}{l}{a}_n=\frac{1}{1+a_{n-1}}\left\{nthterm\right\}\\ a_2=\frac{1}{1+a_{2-1}}\{n=2\}\\ a_2=\frac{1}{1+a_1}\{a_1=1\}\\ a_2=\frac{1}{1+1}\\ a_2=\frac{1}{2}\end{array}$

Next, find $a_3$ ,

  $\begin{array}{l}{a}_n=\frac{1}{1+a_{n-1}}\left\{nthterm\right\}\\ a_3=\frac{1}{1+a_{3-1}}\{n=3\}\\ a_3=\frac{1}{1+a_2}\{a_2=\frac{1}{2}\}\\ a_3=\frac{1}{1+\frac{1}{2}}\\ a_3=\frac{2}{3}\end{array}$

Next, find $a_4$ ,

  $\begin{array}{l}{a}_n=\frac{1}{1+a_{n-1}}\left\{nthterm\right\}\\ a_4=\frac{1}{1+a_{4-1}}\{n=4\}\\ a_4=\frac{1}{1+a_3}\{a_3=\frac{2}{3}\}\\ a_4=\frac{1}{1+\frac{2}{3}}\\ a_4=\frac{3}{5}\end{array}$

Finally find $a_5$ ,

  $\begin{array}{l}{a}_n=\frac{1}{1+a_{n-1}}\left\{nthterm\right\}\\ a_5=\frac{1}{1+a_{5-1}}\{n=5\}\\ a_5=\frac{1}{1+a_4}\{a_4=\frac{3}{5}\}\\ a_5=\frac{1}{1+\frac{3}{5}}\\ a_5=\frac{5}{8}\end{array}$

Conclusion:

The first five term of the sequence are $1,\frac{1}{2},\frac{2}{3},\frac{3}{5},\frac{5}{8}$ .