Solution
= Solution
<Linearity of expectation> gives $\mathbb E(X+Y)=\mathbb EX+\mathbb EY$. Expand the centered product and use <linearity of expectation> again:
$$
\begin{aligned}
\operatorname{cov}(X+Y,Z)
&=\mathbb E\big[((X-\mathbb EX)+(Y-\mathbb EY))(Z-\mathbb EZ)\big]\\
&=\operatorname{cov}(X,Z)+\operatorname{cov}(Y,Z).
\end{aligned}
$$
Thus \b[the assertion is true whenever the covariances are defined]. Neither <independence> nor identical distributions are required.