Question

If $n$ is a positive integer, then $(\sqrt{3}+1{)}^{2n}-(\sqrt{3}-1{)}^{2n}$ is:

• A. an irrational number
• B. an odd positive integer
• C. an even positive integer
• D. a rational number other than positive integers

Answer 1

The correct option is A an irrational number

Given,

$(\sqrt{3}+1{)}^{2n}-(\sqrt{3}-1{)}^{2n}$

The above question can be broken into
$\left[\right((\sqrt{3}+1{)}^2{)}^n-((\sqrt{3}-1{)}^2{)}^n]$$=2^n\left[\right(2+\sqrt{3}{)}^n-(2-\sqrt{3}{)}^n]$

All the terms at the odd places will cancel out and hence we will get
$2^n\left(2\right)\left[2^{n-1}\right(\sqrt{3}{)}^1{}^n{C}_1+2^{n-3}(\sqrt{3}{)}^3{}^n{C}_3+2^{n-5}(\sqrt{3}{)}^5{}^n{C}_5+...]$

Taking $\left(\sqrt{3}\right)$ common we get
$2^{n+1}\left(\sqrt{3}\right)[2^{n-1}{}^n{C}_1+2^{n-3}(\sqrt{3}{)}^2{}^n{C}_3+2^{n-5}(\sqrt{3}{)}^4{}^n{C}_5+...]$$=2^{n+1}\left(\sqrt{3}\right)\left(a\right)$ where $a$ is rational number.

Therefore, the above binomial yields an irrational number.

Hence, option ${}^{\prime }{A}^{\prime }$ is correct. .