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

Question

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

Dr. Abubacker · Accepted Answer

(d)
$\because\ I + f + f^{'} = {(5 + 2\sqrt{6})}^{n} + {(5 - 2\sqrt{6})}^{n}$
$= 2k$(even integer)
$	herefore\ 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 $I = \frac{1}{(1 - f)} - f$

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

Explanation

Reviewed by Dr. Abubacker

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