Differential EquationsHardBoundary Values & Fourier Series ShareBoundary Values & Fourier Series · Hard problemFind the Fourier sine series of f(x)=x(π−x)f(x) = x(\pi - x)f(x)=x(π−x) on [0,π][0, \pi][0,π], and evaluate it at x=π2x = \dfrac\pi2x=2π to find ∑k=0∞(−1)k(2k+1)3=1−127+1125−⋯\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^3} = 1 - \dfrac1{27} + \dfrac1{125} - \cdotsk=0∑∞(2k+1)3(−1)k=1−271+1251−⋯.Check Show answer Walk through it step by step Practice more like this Review this concept