If (2+3)n=I+f where I and n are positive integers and 0<f<1, then (1−f)(I+f)

If (2+3)n=I+f where I and n are positive integers and 0

A
-1
B
1
C
2
D
None of these

Solution:

$I + f = (2 + \sqrt{3})^{n} = 2^{n} +^{n} C_{1} 2^{n - 1} \sqrt{3} +^{n} C_{2} 2^{n - 2} (\sqrt{3})^{2} +^{n} C_{3} 2^{n - 3} (\sqrt{3})^{3} + \dots \dots (i)$

$\begin{array}{ll} Now, 0 < 2 - \sqrt{3} < 1 \\ \Rightarrow 0 < (2 - \sqrt{3})^{n} < 1 \end{array}$

$Let (2 - \sqrt{3})^{n} = f^{'} where 0 < f^{'} < 1$

$\Rightarrow f^{'} = 2^{n} -^{n} C_{1} 2^{n - 1} \sqrt{3} +^{n} C_{2} 2^{n - 2} (\sqrt{3})^{2}$ $-^{n} C_{3} 2^{n - 3} (\sqrt{3})^{3} + \dots \dots (ii)$

On adding Eqs. (i) and (ii), we get

$I + f + f^{'} = 2 [2^{n} +^{n} C_{2} 2^{n - 2} \cdot 3 + \dots]$ -------------(iii)

$\Rightarrow I + f + f^{'} = even integer$

Now $0 < f < 1$

and $0 < f^{'} < 1 \Rightarrow 0 < f + f^{'} < 2$

Hence, from Eq. (iii) we conclude that f + f is an
integer between 0 and 2.

$\Rightarrow f + f^{'} = 1 \Rightarrow f^{'} = 1 - f - - - - (iv)$

$∴ I + f = (2 + \sqrt{3})^{n}, f^{'} = 1 - f = (2 - \sqrt{3})^{n}$

$\Rightarrow (I + f) (1 - f) = [(2 + \sqrt{3}) (2 - \sqrt{3})]^{n} = (4 - 3)^{n} = 1$

$\Rightarrow (I + f) (1 - f) = 1$

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

Solution:

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