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So, I'm having a little difficulty understanding one step in the resolution of an exercise of statistics. In the image below, how did $\sum_{k=a+1}^{b}\frac{b-k+1}{b-a+1}$(1) become $\frac{1}{b-a+1}\sum_{k=1}^{b-a}k$?(2) I understand why $\frac{1}{b-a+1}$ was pulled out of the summation. What I don't understand is where $\sum_{k=1}^{b-a}k$ came from.

$$P(Y\ge k)=\begin{cases}1&k\le a\\\frac{b-k+1}{b-a+1}&a+1\le k\le b\\0&k\ge b+a\end{cases}$$

So

$$\begin{align}\sum_{k=1}^\infty P(y\ge k)&=\sum_{k=1}^a1+\sum_{k=a+1}^b\frac{b-k+1}{b-a+1}\tag1\\&=a+\frac1{b-a+1}\sum_{k=1}^{b-a}k\tag2\\\end{align}$$

(http://i66.tinypic.com/2qx6xqo.png)

Here are the links both for the exercise (3, b) and the solution from which I took the image, just in case someone wants to check it out:

Exercise: https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-fall-2010/assignments/MIT6_041F10_assn03.pdf

Solution: https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-fall-2010/assignments/MIT6_041F10_assn03_sol.pdf

Thanks.

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One trick is to flip the summation backwards:

$$\begin{align}\sum_{k=a+1}^b\underbrace{b-k+1}_{t(k)}&=\underbrace{(b-a)}_{t(a+1)}+\underbrace{(b-a-1)}_{t(a+2)}+\underbrace{(b-a-2)}_{t(a+3)}+\dots+\underbrace{(1)}_{t(b)}\\&=1+2+3+\dots+(b-a)\\&=\sum_{k=1}^{b-a}k\end{align}$$

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    Thanks for your answer, but I'm still a bit lost. When you say, for instance, p(a+1) do you mean the probability that Y equals (a+1)? But shouldn't that equal 1/(b-a+1), since it is a uniform distribution? Thanks – 2017-02-14
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    Ah, horrible choose of letters. I meant to say "expand the sum, and here are the terms in order, then turn them backwards so that we start with the last term and keep adding until we reach the first term, and spot the pattern. – 2017-02-14
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    Oh, now I understand it. One last doubt, if you don't mind: the last term of the last sum would be (b-1), whereas the last term in the expansion above it is (b-a), right? Doesn't it make them different? Thanks for your patience. – 2017-02-14
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    Oh, that is just my bad (minor typos). Good catch! And yes, everything else should be fine. – 2017-02-14