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

∵I+f+f′=(5+26)n+(5−26)n=2k [even integer]∴ f+f′=1Now, (I+f)f′=(5+26)n(5−26)n=(1)n=1⇒ (I+f)(1−f)=1 or I=1(1−f)−f

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

∵I+f+f′=(5+26)n+(5−26)n=2k [even integer]∴ f+f′=1Now, (I+f)f′=(5+26)n(5−26)n=(1)n=1⇒ (I+f)(1−f)=1 or I=1(1−f)−f

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

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$\frac{1}{1 + f} + f$

$\frac{1}{1 - f} - f$

$\frac{1}{f} - f$

$\frac{1}{1 + f} - f$

answer is D.

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$∵ I + f + f^{'} = (5 + 2 \sqrt{6})^{n} + (5 - 2 \sqrt{6})^{n} = 2 k$ [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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If(5+26)n=I+f;n,I∈N and 0≤f<1, then I is equal to

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