First slide

Binomial theorem for positive integral Index

Question

Let $R = (2 + \sqrt{3})^{2 n}$ and $f = R - [R]$ where denotes the greatest integer function, then R(L - f) is equal to

Moderate

A

1
B

2²ⁿ
C

2²ⁿ-1
D

²ⁿC_n

Solution

Given that, $R = (2 + \sqrt{3})^{2 n} and f = R - [R]$
As $0 < 2 - \sqrt{3} < 1, we get 0 < F = (2 - \sqrt{3})^{2 n} < 1$
We have, $\begin{array}{l} R + F = (2 + \sqrt{3})^{2 n} + (2 - \sqrt{3})^{2 n} \\ = 2 [^{2 n} C_{0} 2^{2 n} +^{2 n} C_{2} 2^{2 n - 2} (\sqrt{3})^{2} +^{2 n} C_{4} (2^{2 n - 4}) (\sqrt{3})^{4} + \dots +^{2 n} C_{2 n} (\sqrt{3})^{2 n}] \end{array}$
$\Rightarrow$ R + .F is an even integer.
$\Rightarrow$ [R] + f + F is an even integer.
⇒ f +F is an integer.
But $0 \leq f < 1$ and $0 < F < 1 \Rightarrow 0 < f + F < 2$
But the only integer between 0 and 2 is 1.
Thus,
Now
$\begin{aligned} R (1 - f) = RF = (2 + \sqrt{3})^{2 n} (2 - \sqrt{3})^{2 n} \\ = (4 - 3)^{2 n} = 1^{2 n} = 1 \end{aligned}$

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