First slide

Binomial theorem for positive integral Index

Question

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

Moderate

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

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