Variance of a series of IID’s vs a multiple of a random variable

Question

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?

Answer 1

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}