Q60 / 160 · TS EAMCET 2021 (06 Aug Shift 1) · Mathematics
The period of $\tan k y+\sin k y$, where $k=1+4+9+\ldots 20$ terms, is
- Correct answerA$\frac{\pi}{1435}$
- B$\frac{2 \pi}{1435}$
- C$\pi$
- D$2 \pi$
Solution
$k=1+4+9+\ldots$ upto 20 terms
$$
\begin{aligned}
& \therefore k=\sum n^2=\frac{n(n+1)(2 n+1)}{6}, \text { where } n=20 \\
& \therefore k=\frac{20 \times 21 \times 41}{6}=70 \times 41=2870
\end{aligned}
$$
$\therefore$ Period of tan $k y$ is $\frac{2 \pi}{2870}$
and period of $\sin k y$ is $\frac{2 \pi}{2870}$ i.e. $\frac{\pi}{1435}$
$$
\therefore \text { Required period }=\frac{\pi}{1435}
$$
Hence, the correct answer is option: 'A'.