Let a_k = ( θ_k) i + j and b_k = i - ( θ_k) j , where θ_k = 2^ k-1...

Let $\vec{a}_k = \tan\theta_k\,\hat{i} + \hat{j}$ and $\vec{b}_k = \hat{i} - \cot\theta_k\,\hat{j}$, where $\theta_k = \dfrac{2^{k-1}\pi}{2^n + 1}$.

$|\vec{a}_k|^2 = \tan^2\theta_k + 1 = \sec^2\theta_k$
$|\vec{b}_k|^2 = 1 + \cot^2\theta_k = \csc^2\theta_k$

So the required ratio is:
$\dfrac{\displaystyle\sum_{k=1}^n |\vec{a}_k|^2}{\displaystyle\sum_{k=1}^n |\vec{b}_k|^2} = \dfrac{\displaystyle\sum_{k=1}^n \sec^2\theta_k}{\displaystyle\sum_{k=1}^n \csc^2\theta_k}$

Relation between successive angles:
$\theta_{k+1} = \dfrac{2^k\pi}{2^n + 1} = 2\theta_k$

Key identity:
$\sec^2\theta + \csc^2\theta = \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta} = \dfrac{4}{\sin^2 2\theta} = 4\csc^2 2\theta$

Hence:
$\sec^2\theta = 4\csc^2 2\theta - \csc^2\theta$

Evaluating the numerator using this identity and $2\theta_k = \theta_{k+1}$:
$\displaystyle\sum_{k=1}^n \sec^2\theta_k = \sum_{k=1}^n \left(4\csc^2\theta_{k+1} - \csc^2\theta_k\right) = 4\sum_{k=1}^n \csc^2\theta_{k+1} - \sum_{k=1}^n \csc^2\theta_k$

Let $S_b = \displaystyle\sum_{k=1}^n \csc^2\theta_k$. Then:
$\displaystyle\sum_{k=1}^n \csc^2\theta_{k+1} = S_b - \csc^2\theta_1 + \csc^2\theta_{n+1}$

So:
$\displaystyle\sum_{k=1}^n \sec^2\theta_k = 4\left(S_b - \csc^2\theta_1 + \csc^2\theta_{n+1}\right) - S_b = 3S_b - 4\csc^2\theta_1 + 4\csc^2\theta_{n+1}$

Boundary terms:
$\theta_1 = \dfrac{\pi}{2^n + 1}$, $\theta_{n+1} = \dfrac{2^n\pi}{2^n + 1}$

$\theta_1 + \theta_{n+1} = \dfrac{\pi + 2^n\pi}{2^n + 1} = \dfrac{(2^n + 1)\pi}{2^n + 1} = \pi$

So $\theta_{n+1} = \pi - \theta_1$, and since $\csc(\pi - x) = \csc x$:
$\csc^2\theta_{n+1} = \csc^2\theta_1$

Therefore, $-4\csc^2\theta_1 + 4\csc^2\theta_{n+1} = 0$, and:
$\displaystyle\sum_{k=1}^n \sec^2\theta_k = 3S_b = 3\sum_{k=1}^n \csc^2\theta_k$

Hence, the required ratio is:
$\dfrac{3S_b}{S_b} = 3$

Therefore, the answer is $3$.