Find the $n^{th}$ term of sequence 1, 16, 81, ...

n^{3}

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n^{2}

- 1

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n^{4}

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n^{3}

+ 2n - 2

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Solution

The correct option is C $n^{4}$
$T_{1}$ = $1^{4}$ = 1
$T_{2}$ = $2^{4}$ = 16
$T_{3}$ = $3^{4}$ = 81
thus
$T_{n}$ = $n^{4}$

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