I'm trying to get a better understanding of how to calculate summations and I was trying to solve this question:

$$ \sum_{k=0}^{n-2} (n-1). $$

WolframAlpha gives me an answer of $(n-1)^2$, I already know this is the correct answer from my textbook but the website does not offer any step by step solution to show how it got there, so can someone please explain the steps of this? I just want to better understand this so I can solve bigger questions using this knowledge. All the summation examples I could find online are like from 0 to 5 and they just output some integer at the end.


closed as off-topic by Namaste, mechanodroid, Przemysław Scherwentke, Henrik, Lord Shark the Unknown Oct 13 '17 at 21:14

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  • $\begingroup$ Hint: does the summand depend on $k$? How many terms do you have? $\endgroup$ – Sean Roberson Oct 13 '17 at 18:02

The expression

$$\sum_{k\ = \ 0}^{ n-2}n-1$$

means to add the number $(n-1)$, $(n-1)$ times. That is

$$ \underbrace{(n-1)}_{k=0} + \underbrace{(n-1)}_{k=1} + \cdots + \underbrace{(n-1)}_{k=n-2} = (n-1)(n-1) = (n-1)^2 $$


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