Expected value of total accumulated lifetime (understanding gap in proof) Problem:


I understand the first line $E(T) = ...$
However, I don't get the next two steps. I feel like I almost get it. 
It's like we are factoring out a $\sum_{j=1}^{20}$ but how did he manipulate $\sum_{i=1}^{20}\sum_{j=1}^i$ to get this? 
Does this mean that $\sum_{i=1}^{20}\sum_{j=1}^i = \sum_{j=1}^{20}\sum_{i=j}^{20}$? How does that work?
Also, what's the deal with $\sum_{i=j}^{20}X_j + 30X_j = (51 - j)X_j$?
I can see that the sum $\sum_{i=j}^{20}$ is going to run $20 - j$ times right?
Then wouldn't it be it be $(20 - j)X_j + 30X_j = 20X_j + 30_j - jX_j = (50 - j)X_j$?
Or maybe I need to take a second look at $E(X_i) = \mu/(51 - i)$ from the first slide? 
Also, $X_1$ is exponential with rate $50/\mu$ because you sum all 50 of the $1/\mu$?
confused 
 A: Consider a double sum
$$\sum_{i=1}^{20}\sum_{j=1}^ix_{i,j}\;.$$
You can think of this as being the sum of a triangular array of terms arranged like this:
$$\begin{array}{cc}
x_{1,1}&\\
x_{2,1}&x_{2,2}\\
x_{3,1}&x_{3,2}&x_{3,3}\\
\vdots&\vdots&\vdots&\ddots\\
x_{19,1}&x_{19,2}&x_{19,3}&\dots&x_{19,19}\\
x_{20,1}&x_{20,2}&x_{20,3}&\dots&x_{20,19}&x_{20,20}
\end{array}$$
For a fixed $i$, $\sum_{j=1}^ix_{i,j}$ calculates the sum of the entries in row $i$ of the array, and these row sums are then added by the outer summation to give the total. 
Suppose now that you want to sum the same array first by columns. The row numbers of the entries in column $j$ run from $j$ to through $20$, so the sum of column $j$ is
$$\sum_{i=j}^{20}x_{i,j}\;.\tag{1}$$
To get the sum of all of the entries, just sum $(1)$ over all $20$ columns to get
$$\sum_{j=1}^{20}\sum_{i=j}^{20}x_{i,j}\;.$$
This kind of interchange is a very useful trick: sometimes row sums are easy and column sums are hard, and sometimes it’s the other way around. (Of course sometimes they’re both hard, unfortunately.)
A: The interchanging of the double sum can be seen by writing out the terms with the $\cdots$ notation:
$$
\begin{align}
\sum_{i=1}^{20} \sum_{j=1}^i X_j&=(X_1)+(X_1+X_2)+(X_1+X_2+X_3)+\cdots (X_1+\cdots X_20)\\
&=20\cdot X_1+19\cdot X_2+\cdots+1\cdot X_{20}\\
&=\sum_{j=1}^{20}(20-j+1)\cdot X_j\\
&=\sum_{j=1}^{20} X_j\sum_{i=j}^{20}1\\
&=\sum_{j=1}^{20}\sum_{i=j}^{20}X_j
\end{align}
$$
because $\sum_{i=j}^{20}1=(20-j+1)$ (note that $X_j$ does not depend on $i$).
Also
$$
\sum_{i=j}^{20}X_j +30X_j=(20-j+1)X_j+30X_j=(51-j)X_j.
$$
