How do you write the first five terms of the sequence a_n=(3n^2-n+4)/(2n^2+1)?

Sep 6, 2017

$\left\{2 , \frac{14}{9} , \frac{28}{19} , \frac{48}{33} , \frac{74}{51}\right\}$

Explanation:

We have:

${a}_{n} = \frac{3 {n}^{2} - n + 4}{2 {n}^{2} + 1}$

Assuming that the first term is $n = 1$ as opposed to $n = 0$ then we simply evaluate the expression with $n = 1$, as follows:

Put $n = 1$ then:

${a}_{1} = \frac{3 {\left(1\right)}^{2} - 1 + 4}{2 {\left(1\right)}^{2} + 1}$
$\setminus \setminus \setminus \setminus = \frac{3 - 1 + 4}{2 + 1}$
$\setminus \setminus \setminus \setminus = \frac{6}{3}$
$\setminus \setminus \setminus \setminus = 2$

Similarly Put $n = 2$ then:

${a}_{2} = \frac{3 {\left(2\right)}^{2} - 2 + 4}{2 {\left(2\right)}^{2} + 1}$
$\setminus \setminus \setminus \setminus = \frac{3.4 - 2 + 4}{2.4 + 1}$
$\setminus \setminus \setminus \setminus = \frac{14}{9}$

etc.

The remaining terms have been calculated using excel: