Use indicator random variables to compute the expected value of the sum of $n$ dice

Expectation of a single die $X_i$, is unsurprisingly:

$$ \E[X_k] = \sum_{i=0}^6 i \Pr\{X_k = i\} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5 $$

As for multiple dice:

$$ \E[X] = \E\Big[\sum_{i=1}^nX_i\Big] = \sum_{i=1}^n \E[X_i] = \sum_{i=1}^n 3.5 = 3.5 \cdot n $$