# Covariance of sums of random variables

I need some help understanding an excercise.

Let $X_1, X_2, X_3 \sim N(-2,3).$

(right here there is an ambiguity about the second parameter: is it $\sigma$ or $\sigma^2$ ?)

First they calculate the variance

$$\sigma^2\left(\sum_{i=1}^3 i X_i\right) = \sum_{i=1}^3 i^2 \sigma^2 X_i = 14 \cdot 9 = 126$$

So far I think I understand. This result implies that $\sigma = 3$ . Am I correct?

Then they give the following line without any explanation:

$$\operatorname{cov}\left(\sum_{i=1}^3 i X_i, \sum_{i=1}^3 X_i\right) = \sum_{i=1}^3 i \cdot \sigma^2 X_i = 54$$

Can you explain what happens here? How did they derive $\sum_{i=1}^3 i \cdot \sigma^2 X_i$?

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Imagine expanding the product $(X_1+2X_2+3X_3)(X_1+X_2+X_3)$. By independence, or uncorrelatedness (not mentioned, but necessary) we have $\operatorname{Cov}(X_iX_j)=0$ if $i\ne j$. And of course $\operatorname{Cov}(iX_i,X_i)=i\operatorname{Var}(X_i)$.
Remark: It is distressing that $N(a,b)$ has two interpretations. If the quoted calculation is correct, $b=\sigma$ is the intended interpretation. A gppd solution of the problem is to not use the abbreviation.