Expected value sum of dots

Expected value sum of dots

We throw $n$-times the die. Let $E_n$ be expected value sum of dots (got
in all throws). Compute
$E_1$
$E_2$
$E_3$
$E_4$
So i know how can I do it, for example in 1. I have:
$E_1 = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4
\cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} =
\frac{7}{2}$.
Similar:
$E_2 = 2 \cdot \frac{1}{36} + 3 \cdot \frac{2}{36} + 4 \cdot \frac{3}{36}
+ 5 \cdot \frac{4}{36}+ 6 \cdot \frac{5}{36} + 7 \cdot \frac{6}{36} + 8
\cdot \frac{5}{36} + 9 \cdot \frac{4}{36} + 10 \cdot \frac{3}{36} + 11
\cdot \frac{2}{36} + 12 \cdot \frac{1}{36} = 7$
It is easy but for $E_3$ and $E_4$ it will be compute very long. I suppose
that exist easier way to do this task. I will grateful for your help.