-3
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, mechanodroid, Przemysław Scherwentke, Henrik, Lord Shark the Unknown
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hint: does the summand depend on $k$? How many terms do you have? $\endgroup$ – Sean Roberson Oct 13 '17 at 18:02
1
$\begingroup$

The expression

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

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

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.