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

Explanation

To show that \(\left\lfloor (\sqrt{3} + 1)^{2n} \right\rfloor + 1\) is divisible by \(2^{n+1}\) for \(n \in \mathbb{N}\), we can follow these steps:

### Step 1: Define the expressions
Let:
\[ x = (\sqrt{3} + 1)^{2n} \]
and
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