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

Jan 14, 2018

$- 5 , - 4 , 9 , 40 , 95$

#### Explanation:

Just plug in $n = 1 , 2 , 3 , 4 , 5$ for the formula to get the first five terms.

We know that the $n$th term of the sequence is defined by

${a}_{n} = n \left({n}^{2} - 6\right)$

So,

${a}_{1} = 1 \left(1 - 6\right) = - 5$ (first term)

${a}_{2} = 2 \cdot \left(4 - 6\right) = - 4$ (second term)

${a}_{3} = 3 \left(9 - 6\right) = 9$ (third term)

${a}_{4} = 4 \left(16 - 6\right) = 40$ (fourth term)

${a}_{5} = 5 \left(25 - 6\right) = 95$ (fifth term)

Hence, the first five terms of the sequence ${a}_{n} = n \left({n}^{2} - 6\right)$ are

$- 5 , - 4 , 9 , 40 , 95$

Jan 14, 2018

$- 5 , - 4 , 9 , 40 , 95$

#### Explanation:

$\text{substitute corresponding value for n into } {a}_{n} = n \left({n}^{2} - 6\right)$

${a}_{n} = n \left({n}^{2} - 6\right) = {n}^{3} - 6 n$

$\Rightarrow {a}_{\textcolor{red}{1}} = {\left(\textcolor{red}{1}\right)}^{3} - \left(6 \times \textcolor{red}{1}\right) = 1 - 6 = - 5$

$\Rightarrow {a}_{\textcolor{red}{2}} = {\left(\textcolor{red}{2}\right)}^{3} - \left(6 \times \textcolor{red}{2}\right) = 8 - 12 = - 4$

$\Rightarrow {a}_{\textcolor{red}{3}} = {\left(\textcolor{red}{3}\right)}^{3} - \left(6 \times \textcolor{red}{3}\right) = 27 - 18 = 9$

$\Rightarrow {a}_{\textcolor{red}{4}} = {\left(\textcolor{red}{4}\right)}^{3} - \left(6 \times \textcolor{red}{4}\right) = 64 - 24 = 40$

$\Rightarrow {a}_{\textcolor{red}{5}} = {\left(\textcolor{red}{5}\right)}^{3} - \left(6 \times \textcolor{red}{5}\right) = 125 - 30 = 95$