Let (a+x)n= Odd trems(A) + Even terms (B) So (a−x)n= Odd terms(A) - Even terms(B) ∴(a+x)n−(a−x)n =(A+B)−(A−B) =2B =2[ even terms] =2[T2+T4+T6+.......] So in case of (√3+1)2n−(√3−1)2n =2[T2+T4+T6+......] =2[2nC1⋅(√3)2n−1+2nC3⋅(√3)2n−3+.... Here in (√3)2n−1,2n−1 is odd number. So there will be always √3 in (√3)2n−1. So (√3+1)2n−(√3−1)2n will be always irrational number.
