Chapter 11, Problem 96CR

To determine

To evaluate: the first five terms of the sequence and find the limit of the sequence.

Answer to Problem 96CR

The first five terms of sequence is $5,3,\frac{13}{5},\frac{17}{7}\text{ and }\frac{7}{3}$ .

The limit of sequence is $\infty$ .

Explanation of Solution

Given:

  $a_n=\frac{4n^3+1}{2n-1}$

In this, finding the first term as n = 1,

  $\begin{array}{l}{a}_n=\frac{4n^3+1}{2n-1}\\ a_1=\frac{4\left(1\right)+1}{2\left(1\right)-1}\\ a_1=5\end{array}$

In this, finding the first term as n = 2,

  $\begin{array}{l}{a}_n=\frac{4n^3+1}{2n-1}\\ a_2=\frac{4\left(2\right)+1}{2\left(2\right)-1}\\ a_2=3\end{array}$

In this, finding the first term as n = 3,

  $\begin{array}{l}{a}_n=\frac{4n^3+1}{2n-1}\\ a_3=\frac{4\left(3\right)+1}{2\left(3\right)-1}\\ a_3=\frac{13}{5}\end{array}$

In this, finding the first term as n = 4,

  $\begin{array}{l}{a}_n=\frac{4n^3+1}{2n-1}\\ a_3=\frac{4\left(4\right)+1}{2\left(4\right)-1}\\ a_3=\frac{17}{7}\end{array}$

In this, finding the first term as n = 5,

  $\begin{array}{l}{a}_n=\frac{4n^3+1}{2n-1}\\ a_5=\frac{4\left(5\right)+1}{2\left(5\right)-1}\\ a_5=\frac{21}{9}\\ a_5=\frac{7}{3}\end{array}$

Finding the limit of sequence,

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{4n^2+1}{2n-1}\\ \underset{n\to \infty }{\lim }\frac{4n^2+1}{2n-1}=\underset{n\to \infty }{\lim }\left(\frac{n\left(4n+\frac{1}{n}\right)}{n\left(2-\frac{1}{n}\right)}\right)\\ \underset{n\to \infty }{\lim }\left(\frac{n\left(4n+\frac{1}{n}\right)}{n\left(2-\frac{1}{n}\right)}\right)=\infty \end{array}$

Hence, the first five terms of sequence is $5,3,\frac{13}{5},\frac{17}{7}\text{ and }\frac{7}{3}$ .

The limit of sequence is $\infty$ .