Question
Find the first four terms of the sequence whose $n^{\text {th }}$ terms are given by $a_n=n^3-2$

Answer

$t _{ n }= a _{ n }= n ^3-2$
$a_1=1^3-2=1-2=-1$
$a_2=2^3-2=8-2=6$
$a_3=3^3-2=27-2=25$
$a_4=4^3-2=64-2=62$
$\therefore$ The first four terms are $-1,6,25,62, \ldots \ldots \ldots .$.

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