Differential Equations 1 Question 6
6. The value of $\sum _{k=1}^{13} \frac{1}{\sin \frac{\pi}{4}+\frac{(k-1) \pi}{6} \sin \frac{\pi}{4}+\frac{k \pi}{6}}$ is equal to
(2016 Adv.)
(a) $3-\sqrt{3}$
(b) $2(3-\sqrt{3})$
(c) $2(\sqrt{3}-1)$
(d) $2(2+\sqrt{3})$
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Answer:
Correct Answer: 6. (c)
Solution:
- Here, $\sum _{k=1}^{13} \frac{1}{\sin \frac{\pi}{4}+\frac{(k-1) \pi}{6} \sin \frac{\pi}{4}+\frac{k \pi}{6}}$
Converting into differences, by multiplying and dividing by $\sin \frac{\pi}{4}+\frac{k \pi}{6}-\frac{\pi}{4}+\frac{(k-1) \pi}{6}$, i.e. $\sin \frac{\pi}{6}$.
$$ \therefore \sum _{k=1}^{13} \frac{\sin \frac{\pi}{4}+k \frac{\pi}{6}-\frac{\pi}{4}+(k-1) \frac{\pi}{6}}{\sin \frac{\pi}{6} \sin \frac{\pi}{4}+(k-1) \frac{\pi}{6} \sin \frac{\pi}{4}+k \frac{\pi}{6}} $$
$$ \begin{aligned} & =2 \sum _{k=1}^{13} \frac{\sin \frac{\pi}{4}+\frac{k \pi}{6} \cos \frac{\pi}{4}+(k-1) \frac{\pi}{6}}{\sin \frac{\pi}{4}+(k-1) \frac{\pi}{6} \sin \frac{\pi}{4}+k \frac{\pi}{6}} \\ & =2 \sum _{k=1}^{13} \cot \frac{\pi}{4}+(k-1) \frac{\pi}{6}-\cot \frac{\pi}{4}+k \frac{\pi}{6} \\ & =2 \cot \frac{\pi}{4}-\cot \frac{\pi}{4}+\frac{\pi}{6} \\ & +\quad+\cot \frac{\pi}{4}+\frac{\pi}{6}-\cot \frac{\pi}{4}+\frac{2 \pi}{6} \\ & +2+\cot \frac{\pi}{4}+12 \frac{\pi}{6}-\cot \frac{\pi}{4}+13 \frac{\pi}{6} \\ & =2 \cot \frac{\pi}{4}-\cot \frac{\pi}{4}+13 \frac{\pi}{6} \\ & =21-\cot \frac{29 \pi}{12}=21-\cot 2 \pi+\frac{5 \pi}{12} \\ & =2(1-2+\sqrt{3}) \\ & =2(\sqrt{3}-1) \end{aligned} $$
