Let ak⃗=(tanθk)i^+j^\vec{a_k}=(\tan\theta_k)\hat i+\hat jak=(tanθk)i^+j^ and bk⃗=i^−(cotθk)j^\vec{b_k}=\hat i-(\cot\theta_k)\hat jbk=i^−(cotθk)j^, where θk=2k−1π2n+1\theta_k=\dfrac{2^{k-1}\pi}{2^n+1}θk=2n+12k−1π, for some n∈Nn\in\mathbb Nn∈N, n>5n>5n>5. Then the value of ∑k=1n∣ak⃗∣2∑k=1n∣bk⃗∣2\dfrac{\sum_{k=1}^n\left|\vec{a_k}\right|^2}{\sum_{k=1}^n\left|\vec{b_k}\right|^2}∑k=1nbk2∑k=1n∣ak∣2 is ________.
Official answer: 3
Magnitudes. ∣a⃗k∣2=sec2θk|\vec a_k|^2=\sec^2\theta_k∣ak∣2=sec2θk and ∣b⃗k∣2=csc2θk|\vec b_k|^2=\csc^2\theta_k∣bk∣2=csc2θk, with θk+1=2θk\theta_{k+1}=2\theta_kθk+1=2θk and θn+1=π−θ1\theta_{n+1}=\pi-\theta_1θn+1=π−θ1.
Identity. sec2θ=4csc22θ−csc2θ\sec^2\theta=4\csc^22\theta-\csc^2\thetasec2θ=4csc22θ−csc2θ. With S=∑csc2θkS=\sum\csc^2\theta_kS=∑csc2θk:∑k=1nsec2θk=4∑k=2n+1csc2θk−S=4(S−csc2θ1+csc2θn+1)−S=3S,\sum_{k=1}^n\sec^2\theta_k=4\sum_{k=2}^{n+1}\csc^2\theta_k-S=4\left(S-\csc^2\theta_1+\csc^2\theta_{n+1}\right)-S=3S,k=1∑nsec2θk=4k=2∑n+1csc2θk−S=4(S−csc2θ1+csc2θn+1)−S=3S,since csc2θn+1=csc2θ1\csc^2\theta_{n+1}=\csc^2\theta_1csc2θn+1=csc2θ1.
Ratio 3.
Answer: 3