<s>[INST] 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.
 [/INST] Let $X_n$ denote the number of dots thrown on the $n$th try. Then $$E_n = E[\sum_1^n X_i] = \sum_1^nE[X_i] = n E_1$$
do you see how to finish?
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