Chapter 9, Problem 1RE

To determine

To find the first four terms and the 40^th term of the given sequence.

Answer to Problem 1RE

First four terms and 40^th team is,

  $\frac{-1}{2},\frac{3}{5},\frac{-2}{3},\frac{5}{7}\text{and}\frac{41}{43}$

Explanation of Solution

Given information:

Given sequence is,

  $an={\left(-1\right)}^x.\frac{n+1}{n+3}$ For all $n\ge 1$

CALCULATION:

Given sequence is,

  $an={\left(-1\right)}^x.\frac{n+1}{n+3}$

To find the first term of the above given sequence. Find $a_1$ where $n=1$ .

  $a_1={\left(-1\right)}^1.\frac{1+1}{1+3}=-1.\frac{2}{4}=\frac{-1}{2}\left[\text{since, }n=1\gcd \text{ of '2' and'4' is 2}\right]$

To find the 2^nd term of the above given sequence. Find $a_2$ where $n=2$ .

  $a_2={\left(-1\right)}^2.\frac{2+1}{1+3}=\frac{3}{5}\left[\text{since, }n=2\right]$

To find the third term of the above given sequence. Find $a_3$ where $n=3$ .

  $a_3={\left(-1\right)}^3.\frac{3+1}{3+3}=-\frac{4}{6}=\frac{-2}{3}\left[\text{since, }n=3\gcd \text{ of 4 and6 is '2'}\right]$

To find the fourth term of the above given sequence. Find $a_4$ where $n=4$

  $a_4={\left(-1\right)}^4.\frac{4+1}{4+3}=\frac{5}{7}\left[\text{since, }n=4\right]$

To find the 40^th term of the above given sequence. Find $a_{40}$ where $n=40$

  $a_{40}={\left(-1\right)}^{40}.\frac{40+1}{40+3}=\frac{41}{43}\left[\text{since, }n=4\right]$

Hence,

The first four term and 40^th term is $\frac{-1}{2},\frac{3}{5},\frac{-2}{3},\frac{5}{7}\text{and}\frac{41}{43}$ .