My teacher comes up with an expression saying:

an + (n+1)^{2} = 1/6[n(n + 1)(2n + 1) + 6(n + 1)^{2}]

I read on the internet that [ ] is used for a variaty of things. namingsly intervals, floor, etc. But in this situation what would they mean?

P.S an = n(n + 1)(2n + 1)/6

  • 1
    $\begingroup$ It looks like your teacher is just using them as parentheses - in other words they have the same meaning as '(' and ')', but are used in their place to avoid confusion (since the 'round' parentheses have already been used once) $\endgroup$
    – user259242
    Nov 4, 2016 at 12:43
  • $\begingroup$ Is the right hand side of your equation supposed to be $\frac{1}{6[n(n+1)(2n+1) + 6(n+1)^{2}]}$ or just $\frac{1}{6}[n(n+1)(2n+1) + 6(n+1)^{2}]$? Parentheses are your friend! :) $\endgroup$
    – layman
    Nov 4, 2016 at 12:44
  • $\begingroup$ The brackets are just serving as parentheses here, i.e. grouping terms. $\endgroup$ Nov 4, 2016 at 12:45
  • $\begingroup$ ohh! i see your point user469444! $\endgroup$
    – Nulle
    Nov 4, 2016 at 12:49

2 Answers 2


In the above context [] is the same as ()

  • $\begingroup$ Okay thank you very much. We learn something new every day! $\endgroup$
    – Nulle
    Nov 4, 2016 at 12:47
  • $\begingroup$ Using different styles of parentheses is done some times to make it easier on the eyes of the reader. It's not always easy to parse brackets and tell which closing brackets corresponds which opening bracket. You could start counting brackets, but then you lose sight of the expression as a whole. Different styles is a great help. $\endgroup$
    – Arthur
    Nov 4, 2016 at 12:48

Compare $$\frac16(n(n+1)(2n+1)+6(n+1)^2)$$ with $$\frac16\left[n(n+1)(2n+1)+6(n+1)^2\right]$$

The expression is the same, but the readability and clarity is improved.


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