Verified Solution • 2025–26 Session moderate

QJEEM12Ma038b: JEE Main 2012 Mathematics - Class 11 JEEM12Ma038b Solution: If is a positive integer, then is :

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

Multiple Choice Options

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

Step-by-Step Verified Solution

Let (a+x)n{\left( {a + x} \right)^n} = Odd trems(A) + Even terms(B)

So (a−x)n{\left( {a - x} \right)^n} = Odd terms(A) - Even terms(B)

∴\therefore (a+x)n−(a−x)n{\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}

= (A + B) - (A - B)

= 2B

= 2[even terms]

= 2[ T2 + T4 + T6 + ....... ]

So in case of (3+1)2n−(3−1)2n{\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}

= 2[ T2 + T4 + T6 + ....... ]

= 2[ 2nC1.(3)2n−1+2nC3.(3)2n−3+...{}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...

Here in (3)2n−1{\left( {\sqrt 3 } \right)^{2n - 1}} , 2n - 1 is odd number. So there will be always 3{\sqrt 3 } in (3)2n−1{\left( {\sqrt 3 } \right)^{2n - 1}} .

So (3+1)2n−(3−1)2n{\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}} will be always irrational number.

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