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

`because 3^(2^(n))-1=(4-1)^(2^(n))-1`
`=(4^(2^(n)) -""^(2^(n))C_(1) cdot 4^(2^(n)-1) + ""^(2^(n))C_(2) cdot 4 ^(2^(n)-2)-...-""^(2^(n)) C_(2^(n)-1)cdot4+1)-1`
`=4^(2^(n)) - 2^(n)cdot 4^(2^(n)-1)+(2^(n)(2^(n)-1))/2 cdot 4^(2^(n)-2)-...- 2^(n) cdot 4`
`=2^(n+2) (2^(2^(n+1)-n-2)-2^(2^(n+1)-4)+...-1)=2^(n+2)` (Integer)
Hence, `3^(2^(n)` -1 is divisible by `2^(n+2).lambda = n +2`