Solution verification for this Fourier series problem
You made a small mistake in the final step. The second term in parentheses should be $2((-1)^k-1)$ because you have two terms $e^{-j\pi k}$. So your final result should be
$$X[k]=\frac{3}{2\pi jk}\big(1-(-1)^k\big)\tag{1}$$
With $(-1)^k=e^{-j\pi k}$, Eq. $(1)$ can be rewritten as
$$\begin{align}X[k]&amp;=\frac{3}{k\pi}\frac{1-e^{-jk\pi}}{2j}\\&amp;=\frac{3e^{-jk\pi /2}}{k\pi}\frac{e^{jk\pi /2}-e^{-jk\pi /2}}{2j}\\&amp;=\frac{3e^{-jk\pi /2}\sin(k\pi/2)}{k\pi}\tag{2}\end{align}$$