How do you write the first five terms of the sequence $a_{n} = \frac{6 n}{3 n^{2} - 1}$ $a_n=(6n)/(3n^2-1)$ ?

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Answer 1 · 2017-04-24T07:07:26

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

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

As $a_{n} = \frac{6 n}{3 n^{2} - 1}$ $a_n=(6n)/(3n^2-1)$ ,

$a_{1} = \frac{6 \times 1}{3 \times 1^{2} - 1} = \frac{6}{3 - 1} = \frac{6}{2} = 3$ $a_1=(6xx1)/(3xx1^2-1)=6/(3-1)=6/2=3$

$a_{2} = \frac{6 \times 2}{3 \times 2^{2} - 1} = \frac{12}{12 - 1} = \frac{12}{11}$ $a_2=(6xx2)/(3xx2^2-1)=12/(12-1)=12/11$

$a_{3} = \frac{6 \times 3}{3 \times 3^{2} - 1} = \frac{18}{27 - 1} = \frac{18}{26} = \frac{9}{13}$ $a_3=(6xx3)/(3xx3^2-1)=18/(27-1)=18/26=9/13$

$a_{4} = \frac{6 \times 4}{3 \times 4^{2} - 1} = \frac{24}{48 - 1} = \frac{24}{47}$ $a_4=(6xx4)/(3xx4^2-1)=24/(48-1)=24/47$

$a_{5} = \frac{6 \times 5}{3 \times 5^{2} - 1} = \frac{30}{75 - 1} = \frac{30}{74} = \frac{15}{37}$ $a_5=(6xx5)/(3xx5^2-1)=30/(75-1)=30/74=15/37$

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

How do you write the first five terms of the sequence an=6n3n2−1a_n=(6n)/(3n^2-1)?