Let S_n = ∑_ k=1 ^ n (((2k-1)π)/(12)) (((2k+1)π)/(12)) . If S_n = 2...

The general term of the series is $T_k = \sec\left(\dfrac{(2k-1)\pi}{12}\right) \sec\left(\dfrac{(2k+1)\pi}{12}\right) = \dfrac{1}{\cos\left(\dfrac{(2k-1)\pi}{12}\right) \cos\left(\dfrac{(2k+1)\pi}{12}\right)}$.

Notice that the difference between the angles is $\dfrac{(2k+1)\pi}{12} - \dfrac{(2k-1)\pi}{12} = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$.

Multiplying and dividing by $\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$, we get:

$T_k = \dfrac{2 \sin\left(\dfrac{(2k+1)\pi}{12} - \dfrac{(2k-1)\pi}{12}\right)}{\cos\left(\dfrac{(2k-1)\pi}{12}\right) \cos\left(\dfrac{(2k+1)\pi}{12}\right)} = 2 \left[ \tan\left(\dfrac{(2k+1)\pi}{12}\right) - \tan\left(\dfrac{(2k-1)\pi}{12}\right) \right]$

Summing from $k=1$ to $n$, the series telescopes:

$S_n = \sum_{k=1}^{n} T_k = 2 \left[ \tan\left(\dfrac{3\pi}{12}\right) - \tan\left(\dfrac{\pi}{12}\right) + \tan\left(\dfrac{5\pi}{12}\right) - \tan\left(\dfrac{3\pi}{12}\right) + \dots + \tan\left(\dfrac{(2n+1)\pi}{12}\right) - \tan\left(\dfrac{(2n-1)\pi}{12}\right) \right]$

$S_n = 2 \left[ \tan\left(\dfrac{(2n+1)\pi}{12}\right) - \tan\left(\dfrac{\pi}{12}\right) \right]$

We know that $\tan\left(\dfrac{\pi}{12}\right) = \tan(15^\circ) = 2 - \sqrt{3}$. Substituting this into the expression:

$S_n = 2 \tan\left(\dfrac{(2n+1)\pi}{12}\right) - 2(2 - \sqrt{3}) = 2 \tan\left(\dfrac{(2n+1)\pi}{12}\right) - 4 + 2\sqrt{3}$

We are given that $S_n = 2\sqrt{3} - 6$. Equating the two expressions:

$2 \tan\left(\dfrac{(2n+1)\pi}{12}\right) - 4 + 2\sqrt{3} = 2\sqrt{3} - 6$

$2 \tan\left(\dfrac{(2n+1)\pi}{12}\right) = -2 \Rightarrow \tan\left(\dfrac{(2n+1)\pi}{12}\right) = -1$

The general solution for $\tan x = -1$ is $x = m\pi - \dfrac{\pi}{4}$, where $m \in \mathbb{Z}$.

$\dfrac{(2n+1)\pi}{12} = m\pi - \dfrac{\pi}{4} = \dfrac{(4m-1)\pi}{4} = \dfrac{(12m-3)\pi}{12}$

Equating the numerators: $2n + 1 = 12m - 3 \Rightarrow 2n = 12m - 4 \Rightarrow n = 6m - 2$.

For $n$ to be a positive integer, $m$ must be a positive integer ($m = 1, 2, 3, \dots$).

For $m = 1$, $n = 4$.
For $m = 2$, $n = 10$.
For $m = 3$, $n = 16$.

The sum of the first three positive integer values of $n$ is $4 + 10 + 16 = 30$.

Answer: $30$