$\text{Var}(\sum_{k=1}^n X_i )$ when $X_i$ iid $\text{Var}(\sum_{k=1}^n X_i)$  when $X_k$ iid:
Is it $n\text{Var}(X_k)$ or $n^2 Var(X_k)$? If I write $$\sum_{k=1}^n X_k=nX_k,$$ then it's the second one. If I open the Var using the fact they are independent, I get the first.
I assume the answer is the first, but why is it wrong to write $\sum_{k=1}^n X_k=nX_k$ if they have an identical meaning and distribution?
 A: Let's say that $\text{Var}[X_i] = \sigma^2$ for each $i$. Then we see that
$$\text{Var}\left[X_1+\dotsb+X_n\right] = \text{Var}[X_1]+\dotsb+\text{Var}[X_n]=\sigma^2+\dotsb +\sigma^2 = n[\sigma^2] = n\text{Var}[X_i]$$
where the second step is due to independence. 
As for
$$\sum_{k=1}^n X_i=nX_i,$$
this is not true because this is saying that each realization of $X_i$ is equal to one another, which is not true. For example, if $X_i$ represents the value of a die on the $i$th roll, then it is not true that each $X_i$ is equal to one another necessarily. Specifically, if $n =3$, and we have $X_1 = 1, X_2 = 2, X_3 = 3$, then
$$X_1+X_2+X_3 = 1+2+3 = 5$$
which does not equal $3X_1 = 3$ or $3X_2 = 6$ or $3X_3 = 9$.
A: Think of the example of rolling a 6-sided dice. Your first formula corresponds to rolling it $n$ separate times and taking the sum. Your second one corresponds to rolling it just one time, and then multiplying the result by $n$.
To get a sum of $n$ the first way would be highly unlikely for large $n$: you need to roll a 1 every single time. With the second way, it will happen 1 out of 6 times.
