Question

Period of $\tan (x+4x+9x+..+n^{2}x)$ is

• A. $\frac{2\pi }{n(n+1)(2n+1)}$
• B. $\frac{4\pi }{n(n+1)(2n+1)}$
• C. $\frac{6\pi }{n(n+1)(2n+1)}$
• D. $\frac{8\pi }{n(n+1)(2n+1)}$

Answer 1

Answer: (C)

$\frac{6\pi }{n(n+1)(2n+1)}$

Let $f\left(x\right)=\tan (x+4x+9x....+nx)$
$=\tan \left(\frac{n(n+1)(2n+1)}{6}x\right)$.....( using ${\sum }_{r=0}^n{r}^2=\frac{n(n+1)(2n+1}{6})$

$periodof\tan x=\pi$

$\therefore periodoff\left(x\right)=\frac{6\pi }{n(n+1)(2n+1)}$.................(as if period of $f\left(x\right)=p,periodoff\left(mx\right)=p/m$)