If $(2 + \sqrt{3})^{n} = I + f,$ where $I$ and $n$ are positive integers and $0 < f < 1$ , then the value of $(1 - f) (I + f)$ is

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Solution

$(2 + \sqrt{3})^{n} = I + f I + f = 2^{n} +^{n} C_{1} \cdot 2^{n - 1} \cdot \sqrt{3} +^{n} C_{2} \cdot 2^{n - 2} \cdot (\sqrt{3})^{2} + \dots - (1)$
Now, $0 < (2 - \sqrt{3}) < 1$
Let $(2 - \sqrt{3})^{n} = f^{'}$
$∴ f^{'} = 2^{n} -^{n} C_{1} \cdot 2^{n - 1} \cdot \sqrt{3} +^{n} C_{2} \cdot 2^{n - 2} \cdot (\sqrt{3})^{2} - \dots - (2)$

Adding $(1)$ and $(2)$
$I + f + f^{'} = 2 (2^{n} +^{n} C_{2} \cdot 2^{n - 2} \cdot (\sqrt{3})^{2} + \dots)$
$∴ I + f + f^{'} = 2 m, m \in N - (3)$

Now $0 < f < 1; 0 < f^{'} < 1$
$∴ 0 < f + f^{'} < 2$
From $(3)$ , $f + f^{'}$ will be an integer between $0$ and $2$
$∴ f + f^{'} = 1 f^{'} = 1 - f . . . - (4)$
$∴ (I + f) (1 - f) = (2 + \sqrt{3})^{n} (2 - \sqrt{3})^{n} = 1$

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