Question
Find the first four terms of the sequences whose $n^{\text {th }}$ terms are given by $a_n=2 n^2-6$

Answer

$a_n=2 n^2-6$
$a_1=2(1)^2-6=2-6=-4$
$a_2=2(2)^2-6=8-6=2$
$a_3=2(3)^2-6=18-6=12$
$a_4=2(4)^2-6=32-6=26$
$\therefore$ The first four terms are $-4,2,12,26, \ldots$

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