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(Pardon if this seems a bit beginner, this is my first post in math - trying to improve my knowledge while tackling Project Euler problems)

I'm aware of Sigma notation, but is there a function/name for e.g.

$$ 4 + 3 + 2 + 1 \longrightarrow 10 ,$$

similar to $$4! = 4 \cdot 3 \cdot 2 \cdot 1 ,$$ which uses multiplication?

Edit: I found what I was looking for, but is there a name for this type of summation?


marked as duplicate by Alex M., Namaste algebra-precalculus Oct 31 '18 at 11:02

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  • $\begingroup$ See Faulhaber's formula. $\endgroup$ – Lucian Jan 12 '17 at 11:25
  • $\begingroup$ I like to call it "additorial" or "sumitorial" :) $\endgroup$ – NH. Apr 23 '18 at 14:52
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    $\begingroup$ @AlexM. absolutely not. He's asking for a term, not a proof of an equality. $\endgroup$ – The Great Duck Oct 31 '18 at 3:56
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    $\begingroup$ @AlexM. That's a stupid reason to mark as duplicate though. When something is marked as duplicate the test says "marked as exact duplicate". Has that been revised since I last saw it? $\endgroup$ – The Great Duck Oct 31 '18 at 23:16
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    $\begingroup$ This question is obviously not a duplicate of the question that it is marked as a duplicate of. $\endgroup$ – Greg Schmit Apr 12 at 16:58

The name for

$$ T_n= \sum_{k=1}^n k = 1+2+3+ \dotsb +(n-1)+n = \frac{n(n+1)}{2} = \frac{n^2+n}{2} = {n+1 \choose 2} $$

is the $n$th triangular number. This picture demonstrates the reasoning for the name:

$$T_1=1\qquad T_2=3\qquad T_3=6\qquad T_4=10\qquad T_5=15\qquad T_6=21$$

$\hskip1.7in$ enter image description here

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    $\begingroup$ What does the last notation in brackets mean? Does it have a name? $\endgroup$ – silkfire Mar 29 '16 at 22:50
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    $\begingroup$ @silkfire - It's called a binomial coefficient $\endgroup$ – Russell Thackston Apr 4 '16 at 17:33
  • $\begingroup$ @RussellThackston Thanks! $\endgroup$ – silkfire Apr 4 '16 at 19:37
  • $\begingroup$ How could we adapt this to be used with a known base number ? Let's say we have a base value of 7. Then we need 7+14, the we need 7+14+21 and so forth. Could this be turned into an excel formula ? $\endgroup$ – Overmind Oct 12 '17 at 6:45
  • $\begingroup$ @Overmind: Note that $$7+14+21+\cdots+(7\times n)=7\times(1+2+3+\cdots+n)=7\times T_n$$ So the formula is quite simple :) $\endgroup$ – Zev Chonoles Oct 12 '17 at 7:39

Donald Knuth in The Art of Computer Programming calls the $n$-th triangular number the "termial function", and denotes it

$$n? = 1 + 2 + ... + n = \sum_{k=1}^n k $$

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    $\begingroup$ Wow, really? What volume/page does he define this termin-ology? $\endgroup$ – Niel de Beaudrap Aug 30 '11 at 1:06
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    $\begingroup$ Now that I am home, I have the 3rd edition of volume one, it is on page 48. $\endgroup$ – tlehman Aug 30 '11 at 2:14
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    $\begingroup$ It's not terminal, it's termial. It also doesn't matter why he put it in his books, it is exactly what the questioner was asking about. $\endgroup$ – tlehman Aug 30 '11 at 12:38
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    $\begingroup$ ah, then I managed to misread it several times. Also –– if you will forgive me –– I was somewhat skeptical that Knuth would deign to give this function a name (especially when I thought that name was supposed to be "terminal", which made little sense to me); I wanted to see for myself, and also see why he would do so. $\endgroup$ – Niel de Beaudrap Aug 30 '11 at 12:41
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    $\begingroup$ @Niel: concerning "for pedagical reasons": I'd say, the additive analogon of a "factor" in a multiplication is "summand", so then it should rather be called "summorial" or "summatorial" $\endgroup$ – Gottfried Helms Oct 4 '11 at 6:14

Actually, I've found what I was looking for.

From the wiki on Summation:

enter image description here

  • $\begingroup$ These numbers are also called the triangular numbers. You might think of the triangular numbers as naming a sequence: 1, 3, 6, 10, 15, 21,... But a sequence of integers is really just a function from $\mathbb{N}$ to $\mathbb{Z}$, so the triangular numbers also name the function you've written above. $\endgroup$ – Jonas Kibelbek Aug 29 '11 at 20:59

Not exactly a name, but note that $$ \sum\limits_{k=1}^{n} k= \frac{n(n+1)}{2}={n+1 \choose 2} $$


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