show that `[(sqrt(3) +1)^(2n)] + 1` is divisible by ` 2^(n+1)`
` AA n in N ` , where [.] denote the greatest integer function .

Explanation

Let ` x = (sqrt() + 1)^(2n) = [x] + f ` …(i)
where , ` 0 le f lt 1`
and ` (sqrt(3) -1)^(2n) = f'` …(ii)
whre , ` 0 lt f' lt 1`
On adding Eqs .(i) and (ii), we get
`[x]+ f+ f' = (sqrt(3) + 1)^(2n) + (sqrt(3) - 1)^(2n)`
`= (4 + 2 sqrt(3))^(n)+ (4- 2 sqrt(3))^(n)`
`2^(n){(2+ sqrt(3))^(n)+ (2-sqrt(3))^(n)}`
` 2^(n) *2 {""^(n)C_(0) (2)^(n) + ""^(n)C_(2)(2)^(n-2)(sqrt(3))^(2) + ""^(n)C_(4) (2)^(n-4) (sqrt(3))^(4) + ...}`
` therefore [x] + f + f' = 2^(n+1) k` , where k is an integer ....(iii)
Hence, (f + f') is an integer .
i.e. ` f + f' = 1 [ because 0 lt (f + f') lt 2]`
From Eq .(iii), we get
` [x]+ 1 = 2^(n+1) k`
` rArr [(sqrt(3) + 1)^(2n)] + 1 ` divisible by ` 2^(n+1) AA n in N ` .