If(5+26)n=I+f;n,I∈N and 0≤f<1, then I is equal to

If $(5 + 2 \sqrt{6})^{n} = I + f; n, I \in N and 0 \leq f < 1, then I is$ equal to

A
$\frac{1}{f} - f$
B
$\frac{1}{1 + f} - f$
C
$\frac{1}{1 + f} + f$
D
$\frac{1}{1 - f} - f$

Solution:

$∵ I + f + f^{'} = (5 + 2 \sqrt{6})^{n} + (5 - 2 \sqrt{6})^{n} = 2 k$ [even integer]

$∴ f + f^{'} = 1$

Now, $(I + f) f^{'} = (5 + 2 \sqrt{6})^{n} (5 - 2 \sqrt{6})^{n} = (1)^{n} = 1$

$\Rightarrow (I + f) (1 - f) = 1 or I = \frac{1}{(1 - f)} - f$

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