The largest integer `lambda` such that `2^(lambda)` divides
`3^(2^(n))-1, ninN` is

Options

A. `n-1`
B. n
C. `n+1`
D. `n+2`

Explanation

To find the largest integer \( \lambda \) such that \( 2^{\lambda} \) divides \( 3^{2^n} - 1 \) for \( n \in \mathbb{N} \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression \( 3^{2^n} - 1 \). We can use the identity \( a^2 - b^2 = (a - b)(a + b) \) to factor this expression.

\[ 3^{2^n} - 1 = (3^{2^{n-1}} - 1)(3^{2^{n-1}} + 1) \]
...