Chapter 11, Problem 100CR

To determine

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

Answer to Problem 100CR

The first five terms of the sequence is $a_1=3,a_2=\frac{9}{2},a_3=5,a_2=\frac{21}{4}\text{ and }{a}_5=\frac{27}{5}$ .

The limit of sequence is 6.

Explanation of Solution

Given:

  $a_n=\frac{3}{n}\left\{n+\frac{2}{n}\left[\frac{n\left(n-1\right)}{2}\right]\right\}$

It is given that

  $a_n=\frac{3}{n}\left\{n+\frac{2}{n}\left[\frac{n\left(n-1\right)}{2}\right]\right\}$

Taking n = 1, to find the first term,

  $\begin{array}{l}{a}_n=\frac{3}{n}\left\{n+\frac{2}{n}\left[\frac{n\left(n-1\right)}{2}\right]\right\}\\ a_1=\frac{3}{1}\left\{1+\frac{2}{1}\left[\frac{1\left(1-1\right)}{2}\right]\right\}\\ a_1=3\left\{1+2\left(0\right)\right\}\\ a_1=3\end{array}$

Taking n = 2, to find the second term,

  $\begin{array}{l}{a}_n=\frac{3}{n}\left\{n+\frac{2}{n}\left[\frac{n\left(n-1\right)}{2}\right]\right\}\\ a_2=\frac{3}{2}\left\{2+\frac{2}{2}\left[\frac{2\left(2-1\right)}{2}\right]\right\}\\ a_2=\frac{3}{2}\left\{2+1\left(1\right)\right\}\\ a_2=\frac{9}{2}\end{array}$

Taking n = 3, to find the third term,

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

Taking n = 4, to find the fourth term,

  $\begin{array}{l}{a}_n=\frac{3}{n}\left\{n+\frac{2}{n}\left[\frac{n\left(n-1\right)}{2}\right]\right\}\\ a_4=\frac{3}{4}\left\{4+\frac{2}{4}\left[\frac{4\left(4-1\right)}{2}\right]\right\}\\ a_4=\frac{3}{4}\left\{4+\frac{1}{2}\left(6\right)\right\}\\ a_4=\frac{21}{4}\end{array}$

Taking n = 5, to find the fourth term,

  $\begin{array}{l}{a}_n=\frac{3}{n}\left\{n+\frac{2}{n}\left[\frac{n\left(n-1\right)}{2}\right]\right\}\\ a_5=\frac{3}{5}\left\{5+\frac{2}{5}\left[\frac{5\left(5-1\right)}{2}\right]\right\}\\ a_5=\frac{3}{5}\left\{5+\frac{2}{5}\left(\frac{5\times 4}{2}\right)\right\}\\ a_5=\frac{3}{5}\left\{5+4\right\}\\ a_5=\frac{27}{5}\end{array}$

Now,

Finding limit of the sequence,

  $\begin{array}{c}{a}_n=\frac{3}{n}\left\{n+\frac{2}{n}\left[\frac{n\left(n-1\right)}{2}\right]\right\}\\ \underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{3}{n}\left\{n+\frac{2}{n}\left[\frac{n\left(n-1\right)}{2}\right]\right\}\\ =\underset{n\to \infty }{\lim }\left\{3+\frac{6}{n^2}\left[\frac{n\left(n-1\right)}{2}\right]\right\}\\ =\underset{n\to \infty }{\lim }\left\{3+\frac{6}{n^2}\times n^2\left[\frac{\left(1-\frac{1}{n^2}\right)}{2}\right]\right\}\\ =\underset{n\to \infty }{\lim }\left\{3+6\left[\frac{\left(1-\frac{1}{n^2}\right)}{2}\right]\right\}\\ =\underset{n\to \infty }{\lim }\left\{3+6\left[\frac{\left(1-0\right)}{2}\right]\right\}\\ ==\underset{n\to \infty }{\lim }\left\{3+3\right\}\\ =6\end{array}$

So,

The first five terms of the sequence is $a_1=3,a_2=\frac{9}{2},a_3=5,a_2=\frac{21}{4}\text{ and }{a}_5=\frac{27}{5}$ .

The limit of sequence is 6.