# How do you write the first five terms of the sequence a_n=(6n)/(3n^2-1)?

Apr 24, 2017

First five terms of the sequence are $\left\{3 , \frac{12}{11} , \frac{9}{13} , \frac{24}{47} , \frac{15}{37}\right\}$

#### Explanation:

We can find the first five terms by putting value of $n$ from $1$ to $5$.

As ${a}_{n} = \frac{6 n}{3 {n}^{2} - 1}$,

${a}_{1} = \frac{6 \times 1}{3 \times {1}^{2} - 1} = \frac{6}{3 - 1} = \frac{6}{2} = 3$

${a}_{2} = \frac{6 \times 2}{3 \times {2}^{2} - 1} = \frac{12}{12 - 1} = \frac{12}{11}$

${a}_{3} = \frac{6 \times 3}{3 \times {3}^{2} - 1} = \frac{18}{27 - 1} = \frac{18}{26} = \frac{9}{13}$

${a}_{4} = \frac{6 \times 4}{3 \times {4}^{2} - 1} = \frac{24}{48 - 1} = \frac{24}{47}$

${a}_{5} = \frac{6 \times 5}{3 \times {5}^{2} - 1} = \frac{30}{75 - 1} = \frac{30}{74} = \frac{15}{37}$

Hence, first five terms of the sequence are $\left\{3 , \frac{12}{11} , \frac{9}{13} , \frac{24}{47} , \frac{15}{37}\right\}$