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If $(6 \sqrt{6}+14)^{2 n+1}=P$ , find PF, where F is the fractional part of P
Option: 1
$20^{2 n}$

Option: 2
$20^{2n-1}$

Option: 3
$20^{2 n+1}$

Option: 4
Nore of these

Answers (1)

best_answer

Let $(6 \sqrt{6}+14)^{2 n+1}=P=I+F$

Where I a is positive integer and $0 \leq \mathrm{F}<1$

Clearly $(6 \sqrt{6}-14)^{2 n+1}=F^{\prime}$ where $0<F^{\prime}<1$

Subtracting, we get $2\left[{ }^{2 n+1} C_{1}(6 \sqrt{6})^{2 n} \times 14+{ }^{2 n+1} C_{3}(6 \sqrt{6})^{2 n-2} \times 14^{3}+\ldots \ldots . .\right]$ $=\mathrm{I}+\mathrm{F}-\mathrm{F}^{\prime}$

$\therefore \mathrm{I}+\mathrm{F}-\mathrm{F}^{\prime}=$ an even integer.

But since $0 \leq F<1 and 0<F^{\prime}<1$ $0 \leq F<1$ and $0<F^{\prime}<1$ , the only possibility is that $F-F^{\prime}=0$ or $F=F^{\prime}$

Hence I is an even integer.

Also $P F=P F^{\prime}=(6 \sqrt{6}+14)^{2 n+1}(6 \sqrt{6}-14)^{2 n+1}$

$=(216-196)^{2 n+1}=20^{2 n+1}$ .

Posted by

Deependra Verma

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