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Question: If n is a positive integer, then \({\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \righ...

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

Explanation

Solution

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 pq,q≠0\dfrac{p}{q},q \ne 0 then it is a rational number otherwise it is an irrational number.
Given equation
(3+1)2n−(3−1)2n{\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,
⇒(3+1)2−(3−1)2\Rightarrow {\left( {\sqrt 3 + 1} \right)^2} - {\left( {\sqrt 3 - 1} \right)^2}
Now expand the square according to the property that (a+b)2=a2+b2+2ab,(a−b)2=a2+b2−2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab,{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab so we have,
⇒(3+1)2−(3−1)2=(3+1+23)−(3+1−23)\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,
⇒(3+1)2−(3−1)2=4+23−4+23=43\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 3\sqrt 3 is an irrational number i.e. it cannot be written in the form of pq,q≠0\dfrac{p}{q},q \ne 0.
Hence (3+1)2n−(3−1)2n{\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 pq,q≠0\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, (a+b)2=a2+b2+2ab,(a−b)2=a2+b2−2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab,{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab.