114
$\begingroup$

(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?

$\endgroup$
3
  • 1
    $\begingroup$ See Faulhaber's formula. $\endgroup$
    – Lucian
    Commented Jan 12, 2017 at 11:25
  • 6
    $\begingroup$ I like to call it "additorial" or "sumitorial" :) $\endgroup$
    – NH.
    Commented Apr 23, 2018 at 14:52
  • 13
    $\begingroup$ This question is obviously not a duplicate of the question that it is marked as a duplicate of. $\endgroup$ Commented Apr 12, 2019 at 16:58

5 Answers 5

137
$\begingroup$

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

$\endgroup$
6
  • 3
    $\begingroup$ What does the last notation in brackets mean? Does it have a name? $\endgroup$
    – silkfire
    Commented Mar 29, 2016 at 22:50
  • 7
    $\begingroup$ @silkfire - It's called a binomial coefficient $\endgroup$ Commented Apr 4, 2016 at 17:33
  • $\begingroup$ @RussellThackston Thanks! $\endgroup$
    – silkfire
    Commented Apr 4, 2016 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
    Commented Oct 12, 2017 at 6:45
  • 1
    $\begingroup$ @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 :) $\endgroup$ Commented Oct 12, 2017 at 7:39
49
$\begingroup$

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

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

(Third edition, Volume 1, page 48).

$\endgroup$
7
  • 4
    $\begingroup$ Wow, really? What volume/page does he define this termin-ology? $\endgroup$ Commented Aug 30, 2011 at 1:06
  • 3
    $\begingroup$ Now that I am home, I have the 3rd edition of volume one, it is on page 48. $\endgroup$
    – tlehman
    Commented Aug 30, 2011 at 2:14
  • 5
    $\begingroup$ 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
    Commented Aug 30, 2011 at 12:38
  • 3
    $\begingroup$ 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. $\endgroup$ Commented Aug 30, 2011 at 12:41
  • 8
    $\begingroup$ @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" $\endgroup$ Commented Oct 4, 2011 at 6:14
12
$\begingroup$

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

From the wiki on Summation:

enter image description here

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Aug 29, 2011 at 20:59
5
$\begingroup$

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

$\endgroup$
-1
$\begingroup$

@Overmind: Note that 7+14+21+⋯+(7×n)=7×(1+2+3+⋯+n)=7×Tn So the formula is quite simple :) – Zev Chonoles Oct 12, 2017 at 7:39

I know its old, and i cant comment on the above in the proper section. But i think this is false. Overmind asked for the sum of a sequence (7,14,21,28,...,...). This posted formula wont do that.

Tn of the sequence = 7n.

7x7n can never give the proper number. I know that 7+14=21.

If n=2.

7x7n= 7x(7xn)= 7x(7x2)= 7x14=98. This is not 21...

Or i am missing something (im no expert, just hobbyist)

$\endgroup$
2
  • $\begingroup$ You missed the part where $T_n$ is defined. We define the $n$th triangular number $T_n$ as the sum of the first $n$ positive integers. It's never anything else. Because of this, $T_2 = 3$ (for example) no matter how you intend to use $T_2$. Therefore $7\times T_2 = 21 = 7+14$. $\endgroup$
    – David K
    Commented Aug 12, 2023 at 1:44
  • $\begingroup$ If you’re still unsure why $7\times T_n$ is correct, you can ask your question using the usual question-asking interface, but copy the address of this question into your question so that people can see what you’re asking about. Then delete this answer. If you understand the answer to your question, you could just delete this answer now. Either way should end up with this answer (which obviously isn’t actually an answer) deleted. I thank you in advance, because afterward I won’t be able to comment here again. $\endgroup$
    – David K
    Commented Aug 12, 2023 at 16:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .