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- #1

$$

\sum_{k=1}^{n}e^{ik\theta} = \frac{e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\frac{\theta}{2}}}{2i\sin\frac{\theta}{2}}

$$

I have a hint that says consider the expression $e^{i\left(n+\frac{1}{2}\right)\theta}-e^{i\left(n-\frac{1}{2}\right)\theta}$.

How can I get the second exponential in the numerator to have that expression?