Question

Find the first six terms of the sequence.a = 2n² - 2

Answer 1

We are given the nth term of a sequence:

$a_n=2n^2\text{ - 2}$

We are required to find the first six terms.

For each term, we substitute for n and evaluate.

First term (n =1)

$\begin{gathered} a_1\text{ = 2 }\times(1)^2\text{ - 2} \\ =\text{ 2 -2 } \\ =\text{ 0} \end{gathered}$

Second term (n = 2)

$\begin{gathered} a_2\text{ =2 }\times(2)^2\text{ - 2} \\ =\text{ 2 }\times\text{ 4 -2} \\ =\text{ 8 - 2} \\ =\text{ 6} \end{gathered}$

Third term (n = 3)

$\begin{gathered} a_3\text{ = 2 }\times(3)^2\text{ - 2} \\ =\text{ 2 }\times\text{ 9 - 2} \\ =\text{ 18 - 2} \\ =\text{ 16} \end{gathered}$

Fourth term (n =4)

$\begin{gathered} a_4\text{ = 2}\times(4)^2\text{ - 2} \\ =\text{ 2 }\times\text{ 16 - 2} \\ =\text{ 32 - 2} \\ =\text{ 30} \end{gathered}$

Fifth term ( n = 5)

$\begin{gathered} a_5\text{ = 2 }\times(5)^2\text{ - 2} \\ =\text{ 2 }\times\text{ 25 - 2} \\ =\text{ 50 - 2} \\ =\text{ 48} \end{gathered}$

Sixth term ( n = 6)

$\begin{gathered} a_6\text{ = 2 }\times(6)^2\text{ - 2} \\ =\text{ 2 }\times\text{ 36 - 2} \\ =\text{ 72 - 2} \\ =\text{ 70} \end{gathered}$

Hence, the first six terms of the sequence are : 0, 6, 16, 30, 48 and 70