sequences and series - Basic Summation Rules

I'm not so good with manipulating summations, and recently going over some questions, I had some problems in understanding how some summation related answers were derived. Here are the two problems. It would be great if anyone could explain to me how these work in detail:

First Problem: fair value of a hat-drawing game

Here in this problem, in Yuval Filmus' solution, he turns the term $$\frac{\sum_{t=X}^{100} t}{100}$$ into $$\frac{(100-X+1)(100+X)}{200}$$

I am quite confused as to how he ended up getting this. I had gotten

$$\frac{\sum_{t=X}^{100} t}{100} = \frac{\sum_{t=0}^{100-X} (t + X)}{100} = \frac{(\sum_{t=0}^{100-X}t) + (\sum_{t=0}^{100-X}X)}{100} = \frac{(100 - X + 1)(100-X)(\frac{1}{2}) + (100-X)(X)}{100} = \frac{(100-X)(100-X+1+2X)}{200}=\frac{(100-X)(100+X+1)}{200}$$

Can anyone tell me where I went wrong?

Second Problem: Expected Value of the Difference between 2 Dice

In this problem I just straight up don't understand how the math works. Can anyone explain (or point me to some resources that I could read) how something like this comes about:

$$\sum_{i=1}^n \sum_{j=1}^n \dfrac{|i-j|}{n^2} = 2 \sum_{i=1}^n \sum_{j=1}^i \frac{i-j}{n^2} = \frac{n^2-1}{3n}$$

I neither understand how the summation formula works for absolute value nor how he got rid of the summations at the end.

Thanks!

Answer

In the first problem, $\sum_{t=0}^{100-X}X=(100-X+1)X$: there are $100-X+1$ terms because of the $t=0$ term. Fix this, and you’ll get Yuval’s answer.

In the second, notice first that $|i-j|=0$ when $i=j$, so we can remove those terms from the sum. Next, $|i-j|=|j-i|$, so we need only sum the terms in which $i>j$ and then double the total to account for those in which $ij$ is
$$\frac1{n^2}\sum_{i=2}^n\sum_{j=1}^{i-1}(i-j)\;:\tag{1}$$

the $1/n^2$ is constant and can be pulled out, $i$ must be at least $2$ to leave room for a smaller $j$, and $j$ runs only up to $i-1$, so as to remain less than $i$. (Robert didn’t pull out the constant factor, and he left in the terms in which $i=j$; since they’re $0$, there’s no actual difference.)

Now rewrite the inner sum: as $j$ runs from $1$ to $i-1$, $i-j$ runs from $i-1$ down to $1$, so $$\sum_{j=1}^{i-1}(i-j)=\sum_{k=1}^{i-1}k=\frac12i(i-1)\;.$$ Substitute this into $(1)$ to get $$\frac1{n^2}\sum_{i=2}^n\frac12i(i-1)=\frac1{2n^2}\sum_{i=2}^ni(i-1)=\frac1{2n^2}\sum_{i=1}^{n-1}i(i+1)\;,$$ where the last step is just an index shift. Now break up the sum to get

$$\begin{align*}

\frac1{2n^2}\sum_{i=1}^{n-1}i(i+1)&=\frac1{2n^2}\left(\sum_{i=1}^{n-1}i^2+\sum_{i=1}^{n-1}i\right)\\
&=\frac1{2n^2}\left(\frac16(n-1)n(2n-1)+\frac12n(n-1)\right)\\
&=\frac{n-1}{12n}(2n-1+3)\\
&=\frac{(n-1)(n+1)}{6n}\\
&=\frac{n^2-1}{6n}\;.
\end{align*}$$

Recall that we have to double this to account for the terms with $i