Fourier Series Solution

Fourier Series Term Evaluation:

We evaluate:

2\Big[(-1)^n(6 - n^2\pi^2)/n^3\Big]\sin(nx) \quad \text{for } n = 1, 2, 3.

For $n = 1$:

2\Big[(-1)^1(6 - 1^2\pi^2)/1^3\Big]\sin(x) = -2(6 - \pi^2)\sin(x).

For $n = 2$:

2\Big[(1)(6 - 4\pi^2)/8\Big]\sin(2x) = \frac{6 - 4\pi^2}{4}\sin(2x).

For $n = 3$:

2\Big[(-1)(6 - 9\pi^2)/27\Big]\sin(3x) = -\frac{6 - 9\pi^2}{27}\sin(3x).

$$-2(6 - \pi^2)\sin(x) \;+\; \frac{6 - 4\pi^2}{4}\sin(2x) \;-\; \frac{6 - 9\pi^2}{27}\sin(3x) + \dots$$

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