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I'm dealing with something like $\sum_{j=2}^{n-1}j^2$. I know I can do this $\sum_{j=1}^{n-1}j^2 - \sum_{j=1}^{1}j^2$.

Would that be equal to $\frac{j(j+1)(2j+1)}{6} - j^2$ or I'm missing some properties with $n-1$?

If so, which ones?

Shaun
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Alex
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Mar 06 '20 at 03:41
  • @Shaun No, I'm just interested in the $n-1$ on top of the summation, how to convert it to just $n$ or if it's okay to leave it like that. – Alex Mar 06 '20 at 04:09
  • After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun Mar 06 '20 at 04:29

2 Answers2

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$$\sum_{j=2}^{n-1}j^2=\sum_{j=1}^{n}j^2-1^2-n^2=\frac{n(n+1)(2n+1)}{6}-1-n^2$$

RobPratt
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Since

$$\sum_{r=1}^mr^2=\frac{m(m+1)(2m+1)}{6},$$

by letting $\color{blue}{m=n-1}$, we have,

$$\begin{align} \sum_{j=2}^{n-1}j^2&=\sum_{j=1}^{n-1}j^2-\color{green}{\sum_{j=1}^1j^2}\\ &=\frac{\color{blue}{(n-1)}(\color{blue}{(n-1)}+1)(2\color{blue}{(n-1)}+1)}{6}-\color{green}{1^2}\\ &=\frac{(n-1)n(2n-1)}{6}-\color{green}{1}. \end{align}$$

Shaun
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