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Question

$4^{3} + 8^{3} + 12^{3} + . . . + n$ terms

{4 n (n + 1)}^{2}

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{8 n (n + 1)}^{2}

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{2 n (n + 1)}^{2}

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{16 n (n + 1)}^{2}

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Solution

The correct option is A ${4 n (n + 1)}^{2}$
$4^{3} + 8^{3} + 12^{3} . . . . . . (4 n)^{3} = 4^{3} (1^{3} + 2^{3} + 3^{3} . . . . . . n^{3})$
$= 4^{3} \sum_{k = 1}^{n} k^{3}$
$= 64 \times \frac{n^{2} (n + 1)^{2}}{4}$
$= 16 \times n^{2} (n + 1)^{2}$
$= (4 n (n + 1))^{2}$

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