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.

If you take $X=X_{1}+,...,+X_{n}$ where $X_{i}, i \in \{1,...,n\}$ is the number of dots on the throw $i$ then you need $E[X]$. So, because expectation of a sum is the sum of the expectations
$$E[X]=E\left[\sum\limits_{i=1}^{n}X_{i}\right]=\sum\limits_{i=1}^{n}E{\left[X_i\right]}$$
but $E[X_i]=E[X_j]$ for all $i,j$ because $X_i$ and $X_j$ are identically distributed and
$$E[X_i]=\sum\limits_{j=1}^{6}j\frac{1}{6}=\frac{1}{6}\frac{42}{2}=\frac{7}{2}$$
then
$$E[X]=\sum\limits_{i=1}^{n}\frac{7}{2}=\frac{7}{2}n.$$
