Solution verification for this Fourier series problem We have a signal with period $T = 2$

We want to find the continuous time fourier series for this signal.
Since $T = 2$, $\omega = \pi$. All we have to do know is find the frequency domain.
$$x(t) = \sum_{k = -\infty}^{\infty}X[k]e^{-i \omega tk}$$
$$X[k] = \frac{1}{T}\int _{T}x(t)e^{i \pi tk}$$
$$X[k] = \frac{1}{2}\int _{2}x(t)e^{-i \pi tk} = \frac{1}{2} \left( \int _{0}^{1}\frac{3}{2}e^{-i \pi tk}dt  + \int _{1}^{2}\frac{-3}{2}e^{-i \pi tk}  dt \right)$$
$$X[k] = \frac{3}{4} \left( \int _{0}^{1}e^{-i \pi tk}dt  - \int _{1}^{2}e^{-i \pi tk}  dt \right) = \frac{3}{4} \left( \frac{-1}{i \pi k} \right)\left( e^{-i \pi k} - 1  - e^{-2i \pi k} + e^{-i \pi k}
\right)$$
$$X[k] = \frac{3}{4} \left( \frac{-1}{i \pi k} \right)\left( (-1)^k - 2\right) \text{(Because k can only be an integer)}$$
This to me is correct... however the answer is actually
$$X[k] = \frac{e^{-j\frac{k\pi}{2}}3\sin(\frac{k\pi}{2})}{k\pi}$$
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
$$\begin{align}X[k]&amp;=\frac{3}{k\pi}\frac{1-e^{-jk\pi}}{2j}\\&amp;=\frac{3e^{-jk\pi /2}}{k\pi}\frac{e^{jk\pi /2}-e^{-jk\pi /2}}{2j}\\&amp;=\frac{3e^{-jk\pi /2}\sin(k\pi/2)}{k\pi}\tag{2}\end{align}$$