The period of tan(ky) + sin(ky),where k = 1 + | vedclass

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(B) The value of $k$ is the sum of the squares of the first $20$ natural numbers:$k = \sum_{n=1}^{20} n^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=20$.$k = \f

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$\frac{2\pi}{1435}$

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(B) The value of $k$ is the sum of the squares of the first $20$ natural numbers:$k = \sum_{n=1}^{20} n^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=20$.$k = \frac{20 	imes 21 	imes 41}{6} = 70 	imes 41 = 2870$.We need the period of $f(y) = 	an(2870y) + \sin(2870y)$.The period of $	an(ky)$ is $T_1 = \frac{\pi}{k} = \frac{\pi}{2870}$.The period of $\sin(ky)$ is $T_2 = \frac{2\pi}{k} = \frac{2\pi}{2870} = \frac{\pi}{1435}$.The period of the sum of two functions is the Least Common Multiple $(LCM)$ of their individual periods.$LCM\left(\frac{\pi}{2870}, \frac{2\pi}{2870}ight) = \frac{LCM(\pi, 2\pi)}{GCD(2870, 2870)} = \frac{2\pi}{2870} = \frac{\pi}{1435}$.

The period of $\tan(ky) + \sin(ky)$,where $k = 1 + 4 + 9 + \ldots$ ($20$ terms),is

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