f(x) =(-x) , if -k<x<0  and f(x) = x  if 0<x<k   $f(-x) = x$ if -k<-x<0  i.e by multiplying with -1 and reversing the order it becomes 0<x<k $f(-x) = -x$ if 0<-x<k  by multiplying -1  -k<x<0 which means f(x) = f(-x)   Hence f(x)  even function   <strong>$\because f(x), is, even function, a_0 = \dfrac{1}{k} \int_{0}^{k} f(x).dx \\ a_n = \dfrac{2}{k}\int_{0}^{k} f(x) .cos(\dfrac{n\pi x}{k}).dx, b_n =0 $     $a_0 =\dfrac{1}{k} \int_{0}^{k} x,dx = \dfrac{1}{k} [\dfrac{x^2}{2}]_0^k = \dfrac{k}{2}$   $a_n =\dfrac{2}{k}\int_{0}^{k} .x(cos(\dfrac{n\pi x}{k}).dx$     $integration ,by , parts\\ a_n = \dfrac{2}{k} [x.sin(\dfrac{\pi nx}{k}) -(-cos(\dfrac{\pi nx}{k}). \dfrac{k^2}{\pi^2.n^2}]_0^k$       $= \dfrac{2}{k}[ksin\pi n .\dfrac{ k}{\pi n} +(cos{\pi n})\dfrac{k^2}{\pi^2n^2}]- 0- \dfrac{k^2}{\pi^2n^2}] $   please note $con\pi n =(-1)^n and , sin\pi n =0$   $a_n=- \dfrac{2k}{\pi^2n^2}(1- cos(-1)^n)$

$f(x) = a_0 + \sum a_n. cos(\dfrac{n\pi x}{k})$

$=\dfrac{k}{2} + \sum \dfrac{-2k}{\pi^2 n^2}[1-(-1)^2].cos(\dfrac{\pi nx}{k})$

$=\dfrac{k}{2}-[\dfrac{2k}{\pi^2 } (2\dfrac{cos\pi x}{k} + 2cos\dfrac{3\pi x}{k}....... ]$

$f(x)= \dfrac{k}{2} -\dfrac{4k}{\pi^2}\sum cos[\dfrac{(2n-1) \pi x}{k}]$

now apply Parseval's identity

NOW APPLY PARSEVAL"S IDENTITY

$if, f(x) = a_0 + \sum a_n cos(\dfrac{n\pi x}{l}) \\ then\ \dfrac{1}{l} f(x)^2 .dx = a_0^2 +\dfrac{1}{2} [ a_1^2 +a_2^2 +a_3^2.......\infty] $

$\dfrac{1}{k}\int_{0}^{k} (x)^2 = \dfrac{K^2} {4} +\dfrac{1}{2}[ \dfrac{4^2k^2}{\pi^4.1^4} +\dfrac{4^2 K^2}{\pi^4 3^4} ......]$

$\dfrac{k^2}{3} = \dfrac{k^2}{4} + 8\dfrac{K^2}{\pi^4}[ \dfrac{1}{1^4} + \dfrac{1}{3^4}............ ]$

$\dfrac{1}{12} = \dfrac{8}{\pi^4}[ \sum \dfrac{1}{(2n-1)^4}]\\ \sum \dfrac{1}{(2n-1)^4}]= \dfrac{\pi^4}{96} deduced $