Summing Gradually Changing Numbers

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The discussion focuses on the concept of triangular numbers, which are generated by summing consecutive integers starting from zero. The sequence includes numbers like 1, 3, 6, 10, and 15, with the nth triangular number calculated using the formula n(n+1)/2. Participants clarify that this summing process is a specific case of an arithmetic sequence, where each term increases by a fixed difference. The average of the first and last terms multiplied by the number of terms yields the sum of the sequence. Overall, the conversation provides insights into identifying triangular numbers and understanding the underlying arithmetic principles.
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Hi.

This is the first Topic I've posted at (the very cool) Physics Forums site so go easy on me. My English is very far from perfect so some used terms may not be used correctly... :)

Anyways... I was writing a script for a website and a mathematics question popped up.

The numbers 1, 3, 6, 10, 15, 21... can be received by summing group of numbers where the first is 0 and each subsequent number is greater then the previous with 1.

for example:

1 = 0 + 1
3 = 0 + 1 + 2
6 = 0 + 1 + 2 + 3
10 = 0 + 1 + 2 + 3 + 4

and so on...

My 2 questions:

(0) I'd love to know if it possible to determine if a given number is among the list of numbers that can be received from such summing.

(1) And if the above is too complicated to discuss here - can you tell me the term in mathematics that describes such summing (if such exists) so I can google the topic?

Any help would be fantastic.

Cheers.
 
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Perfect. Thank You.
 
That's an example of the more general "arithmetic sequence" in which each number is the previous number plus a fixed difference: an+1= an+ d or a1, a1+ d, a1+ 2d, etc. In your example a1= 1 and d= 1.
It can be shown that the sum of a finite sequence of that type is just the average of the first and last terms in the sequence time the number of terms in the sequence. In your case, if you have n terms then the first number is 1 and the last term is n. Their average is (n+1)/2 and so their sum is n(n+1)/2.
 
You're not Gauss, that's for sure :-p
 
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