Write the first five terms of the sequence whose $n^{t h}$ term is $a_{n} = 2^{n}$

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Solution

$a_{n} = 2^{n}$
substituting $n = 1, 2, 3, 4, 5$ we obtain
$a_{1} = 2^{1} = 2$
$a_{2} = 2^{2} = 4$
$a_{3} = 2^{3} = 8$
$a_{4} = 2^{4} = 16$
$a_{5} = 2^{5} = 32$
Therefore the required terms are $2, 4, 8, 16$ and $32$

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