How is this summation simplified? I'm taking EdX's probability class and one question asked to find the expectation of a uniform random variable (problem described here: https://youtu.be/vB6EKsX12hc).
There's one part of the calculation that I don't understand:

how does the numerator in the summation get reduced to a nice and simple k? Where did b+1go? 

$\sum_{i=1}^a 1 = a$.  Sot there's that.
So we just need to show $\sum_{k=a + 1}^b \frac {b-k +1}{b-a - 1} = \frac{1}{b -a - 1}\sum_{k=1}^{b-a} k$.
Which is   $\sum_{k=a + 1}^b \frac {b-k +1}{b-a - 1} =\frac{1}{b -a - 1}\sum_{k=a + 1}^{b} (b - k +1) = \frac{1}{b -a - 1}\sum_{k=1}^{b-a} (b - a - k +1) =\frac{1}{b -a - 1}\sum_{k=1}^{b-a} (b - a - k +1)$
Now for every $1 \le k \le b-a$ there is a unique $i; k = b -a + 1 - i; 1 \le i \le b - a$.
So $\sum_{k=1}^{b-a} (b - a - k +1)= \sum_{i=1}^{b-a} i = \sum_{k=1}^{b-a} k$ so  $\sum_{k=a + 1}^b \frac {b-k +1}{b-a - 1} =\frac{1}{b -a - 1}\sum_{k=a + 1}^{b} (b - k +1) = \frac{1}{b -a - 1}\sum_{k=1}^{b-a} (b - a - k +1) =\frac{1}{b -a - 1}\sum_{k=1}^{b-a} k $
