Binomial Theorem And Its Simple Applications Question 10
Question: If $ x={{( 2+\sqrt{3} )}^{n}} $ , then find the value of $ x,( 1-{ x } ) $ where {x} denotes the fractional part of x
Options:
A) 1
B) 2
C) $ 2^{2n} $
D) $ 2^{n} $
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Answer:
Correct Answer: A
Solution:
- [a] $ x={{( 2+\sqrt{3} )}^{n}} $
$ ={{,}^{n}}C_02^{n}+{{,}^{n}}C_1{2^{n-1}}\sqrt{3}+{{,}^{n}}C_2{2^{n-1}}{{( \sqrt{3} )}^{2}}+… $
Let $ x_1={{( 2-\sqrt{3} )}^{n}} $
$ ={{,}^{n}}C_02^{n}-{{,}^{n}}C_1{2^{n-1}}\sqrt{3}+{{,}^{n}}C_2{2^{n-2}}{{( \sqrt{3} )}^{2}}+… $
$ x+x_1=2( ^{n}C_02^{n}+{{,}^{n}}C_2{2^{n-2}}{{( \sqrt{3} )}^{2}}+… ) $
= Even integer.
Clearly, $ x,\in (0,1)\forall n\in N $
$ \Rightarrow [x]+{x}+x_1= $ Even integer
$ \Rightarrow {x}+x_1=Integer{x}\in (0,1),x_1\in (0,1) $
$ \Rightarrow {x}+x_1\in (0,2) $
$ \Rightarrow {x}+x_1=1\Rightarrow x_1=1-{x} $
$ \Rightarrow x(1-{x})=x.x_1={{( 2+\sqrt{3} )}^{n}}{{( 2-\sqrt{3} )}^{n}}=1 $
BETA
