Let be a dice with the sides $\{0,0,6,0,3,3\}$ which will be rolled $n$ times since $n\in\mathbb{N}$. Now I'm modelling the following experiment: Let $X_i$ be the random variable of the $i$-th throw and $S:=\sum_{i=1}^n X_i$ the sum of the diced values. The exercise is to determine the variance of $S$.
I've calculated the mean of $S$: Since the random variables are independent, then
$$E(X_1) = 0\cdot P(X_1=0)+3\cdot P(X_1=3)+6\cdot P(X_1=6) = 0 + \frac{3}{3}+\frac{6}{6} = 1+1=2,$$ so
$$E(S) = E(X_1+\ldots+X_n) = n\cdot E(X_1) = 2n.$$
Then, the second moment of $X_1$ is
$$E(X_1^2) = 0 + \frac{1}{3}\cdot 9+\frac{1}{6}\cdot 6^2 = 3+6=9,$$
and the variance of $X_1$ is $Var(X_1) = E(X_1^2)-E(X_1)^2 = 9 - 4 = 5.$
For the sum $S$, I determine by the aid of the independece of $X_i$:
$\begin{align*} Var(S) &= Var(X_1+\ldots + X_n) = n\cdot Var(X_1) = 5n.\\ \end{align*}$
Are my calculations and reasons correctly?
All is fine but $$ \operatorname{Var}(X_1+\ldots + X_n) = \operatorname{Var}(n\cdot X_1). $$
You have to take the variance summandwise in case of independence, i.e. $$ \operatorname{Var}(S)= \operatorname{Var}(X_1+\ldots + X_n) =n\cdot \operatorname{Var}(X_1). $$
Illustration: Let $X=\text{weight of a pack of sugar in kg}$. In that case $X_1+\dots +X_n$ is the sum of the weights of onethousand packages in kg whereas $1000 X$ is the weight of one package, measured in grams.