Chapter 12, Problem 1CT

To determine

The first five terms of the sequence $\left\{s_n\right\}=\left\{\frac{n^2-1}{n+8}\right\}$.

Answer to Problem 1CT

Solution:

The first five terms of the sequence $\left\{s_n\right\}=\left\{\frac{n^2-1}{n+8}\right\}$ are $0,\frac{3}{10},\frac{8}{11},\frac{5}{4}$ and $\frac{24}{13}$.

Explanation of Solution

Given information:

The sequence, $\left\{s_n\right\}=\left\{\frac{n^2-1}{n+8}\right\}$.

Explanation:

Consider, $\left\{s_n\right\}=\left\{\frac{n^2-1}{n+8}\right\}$.

First term of the sequence is calculated by substituting $n=1$ in the given sequence.

Thus, $a_1=\frac{1^2-1}{1+8}=\frac{1-1}{9}=0$

Second term of the sequence is calculated by substituting $n=2$ in the given sequence.

$a_2=\frac{2^2-1}{2+8}=\frac{4-1}{10}=\frac{3}{10}$

Third term of the sequence is calculated by substituting $n=3$ in the given sequence.

$a_3=\frac{3^2-1}{3+8}=\frac{9-1}{11}=\frac{8}{11}$

Fourth term of the sequence is calculated by substituting $n=4$ in the given sequence.

$a_4=\frac{4^2-1}{4+8}=\frac{16-1}{12}=\frac{15}{12}=\frac{5}{4}$

Fifth term of the sequence is calculated by substituting $n=5$ in the given sequence.

$a_5=\frac{5^2-1}{5+8}=\frac{25-1}{13}=\frac{24}{13}$

The first five terms of the sequence $\left\{s_n\right\}=\left\{\frac{n^2-1}{n+8}\right\}$ are $0,\frac{3}{10},\frac{8}{11},\frac{5}{4}$ and $\frac{24}{13}$.