Question

If n is a positive integer, then ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$ is
$\left( a \right)$ An irrational number
$\left( b \right)$ An odd positive number
$\left( c \right)$ An even positive number
$\left( d \right)$ A rational number other than positive integers.

Answer 1

Hint: In this particular question assume any least positive integer as n and substitute in the given equation and simplify so use these concepts to reach the solution of the question.

Complete step-by-step answer:
As we all know that if a number is written in the form of $\dfrac{p}{q},q \ne 0$ then it is a rational number otherwise it is an irrational number.
Given equation
${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$, where n is a positive integer.
Let n = smallest positive integer = 1, as 0 is neither considered as positive nor negative so smallest positive integer is 1.
Now substitute n = 1, in the given equation we have,
$ \Rightarrow {\left( {\sqrt 3 + 1} \right)^2} - {\left( {\sqrt 3 - 1} \right)^2}$
Now expand the square according to the property that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab,{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so we have,
$ \Rightarrow {\left( {\sqrt 3 + 1} \right)^2} - {\left( {\sqrt 3 - 1} \right)^2} = \left( {3 + 1 + 2\sqrt 3 } \right) - \left( {3 + 1 - 2\sqrt 3 } \right)$
Now simplify it we have,
$ \Rightarrow {\left( {\sqrt 3 + 1} \right)^2} - {\left( {\sqrt 3 - 1} \right)^2} = 4 + 2\sqrt 3 - 4 + 2\sqrt 3 = 4\sqrt 3 $
Now as we know that $\sqrt 3 $ is an irrational number i.e. it cannot be written in the form of $\dfrac{p}{q},q \ne 0$.
Hence ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$ is an irrational number.
So this is the required answer.
Hence option (a) is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that if a number is written in the form of $\dfrac{p}{q},q \ne 0$ then it is a rational number otherwise it is an irrational number and always recall some of the basic identity such as, ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab,{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.