Variance of a series of IID's vs a multiple of a random variable For any random variable, X, Var(aX) = a^2*Var(X), which is easy to demonstrate.
Suppose you have a series of IID's, and want to find the variance.  So, in that case for example, Var(X+X+X+X+X) = Var(X) + Var(X) + Var(X) + Var(X) + Var(X) = 5Var(X) since there is no covariance involved.  But isn't Var(X+X+X+X+X) = Var(5X) = 25Var(X)?
Follow-up, when doing the variance of a sum of dependent random variables would you add two times every possible pairwise covariance to the individual variances?
 A: If $X_1,...,X_5$ are independent, then
$$V(X_1+X_2+X_3+X_4+X_5) = V(X_1)+V(X_2)+V(X_3)+V(X_4)+V(X_5)$$
For obvious reasons, $X$ is not independent from $X$, so it is your second formula that is correct:
$$V(X+X+X+X+X)=V(5X)=25V(X)$$
A: 
Follow-up, when doing the variance of a sum of dependent random
variables would you add two times every possible pairwise covariance
to the individual variances?

Yes. And the covariance of $X$ and $X$ is the variance: $\operatorname{Cov}(X,X)=\operatorname{Var}(X)$, where $X$ fully depend on itself. That´s why
\begin{align}
\operatorname{Var}(X+X)& =\operatorname{Var}(X)+\operatorname{Var}(X) + 2\cdot \operatorname{Cov}(X,X) \\[6pt]
&=2\cdot \operatorname{Var}(X) + 2 \cdot \operatorname{Var}(X) \\[6pt]
&=4\cdot \operatorname{Var}(X)
\end{align}
A: $\newcommand{\v}{\operatorname{var}}\newcommand{\c}{\operatorname{cov}}$

would you add two times every possible pairwise covariance

Here is a way to think about that. First, suppose you know that
$$
\v(X+Y) = \v(X) + \v(Y) + 2\c(X,Y).
$$
That can be applied repeatedly, as follows:
\begin{align}
& \v(X_1+X_2+X_3+X_4+X_5) \\[6pt]
= {} & \v(X_1+X_2) + \v(X_3+X_4+X_5) \tag1 \\
& {} + 2\c(X_1+X_2,\,X_3+X_4+X_5) \tag2 \\[10pt]
& \text{The first term on line $(1)$ is:} \\[8pt]
& \v(X_1+X_2) = \v(X_1)+\v(X_2) + 2\c(X_1,X_2). \\[8pt]
& \text{The second term on line $(1)$ is:} \\[8pt]
& \v(X_3+X_4+ X_5) \\[4pt]
= {} & \v(X_3)+ \v(X_4+X_5) \\
& {} + 2\c(X_3,\,X_4+X_5). \\[10pt]
& \text{The thing on line $(2)$ is:} \\[8pt]
& \c(X_1+X_2,\,\,X_3+X_4+X_5) \\[8pt]
= {} & \c(X_1,\,\,X_3+X_4+X_5) + \c(X_2,\,\,X_3+X_4+X_5) \tag3 \\[8pt]
= {} & \c(X_1,X_3) + \c(X_1,X_4) + \cdots
\end{align}
Keep going like that until you see just how many covariances you need, and which ones.
