Q01.5MCQMarks unknownFourier series of the periodic function (period 2π) defined by f(x)={0,−p<x<0x,0<x<p is π4+∑[1πn2(cosnπ−1)cosnx−1ncosnπsinnx]f(x) = \begin{cases} 0, -p < x < 0\\x, 0 < x < p \end{cases} \text { is }\\ \frac{\pi}{4} + \sum \left [ \frac{1}{\pi n^2} \left(\cos n\pi - 1 \right) \cos nx - \frac{1}{n} \cos n\pi \sin nx \right ]f(x)={0,−p<x<0x,0<x<p is 4π+∑[πn21(cosnπ−1)cosnx−n1cosnπsinnx] But putting x=πx = \pix=π, we get the sum of the series 1+132+152+172+⋯ is 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots \text { is }1+321+521+721+⋯ is (A)π24\frac{{\pi }^2 }{4}4π2(B)π26\frac{{\pi }^2 }{6}6π2(C)π28\frac{{\pi }^2 }{8}8π2(D)π212\frac{{\pi }^2 }{12}12π2Sign in to save attempts.Show answer← PreviousNext →SolutionsDiscussionLoading…Sign in with Google