If ${(6 \sqrt{6} + 14)}^{n + 1} = m$ and if $f$ is the fractional part of $m$ , then mf is equal to

15^{n + 1}

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20^{n + 1}

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25^{n}

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None of these

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Solution

The correct option is B $20^{n + 1}$
$m f = {(6 \sqrt{6} + 14)}^{n + 1} {(6 \sqrt{6} - 14)}^{n + 1}$
$= {(20)}^{n + 1}$
Hence $,$ Option $(B)$ is correct

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