Find the period of $f (x) = tan (x + 4 x + 9 x + . . . . . . . + n^{2} x)$

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Solution

$f (x) = tan (x + 4 x + 9 x + . . . . . . . + n^{2} x)$
$= tan [x (1 + 4 + 9.... n^{2})]$
$= tan [x \frac{(n) (n + 1) (2 n + 1)}{6}]$
For $tan (x)$ the period is $π$ , so if we multiply $x$ by some value A, the period gets divided by same value A. So, period of $f (x)$ will be
$= \frac{6 π}{n (n + 1) (2 n + 1)}$

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