Hi, can anyone tell me why

1+2+3+...+n=n(n+1)/2

I can see that it works when I choose a number for n, but I don't really see how I could have come up with it myself.

Results 1 to 9 of 9

- September 21st 2011, 02:14 AM #1

- Joined
- Sep 2010
- From
- Oslo
- Posts
- 57
- Thanks
- 4

- September 21st 2011, 02:22 AM #2
## Re: 1+2+3+...+n

I think I can explain it like that:

1, 2, 3, 4, 5, 6, ... n

This is an arithmetic progression with first term 1, last term n with a common difference of 1.

The formula for the sum of the first n numbers is given by:

a = 1, l = n so you simplify to get:

How? Let's take an example:

1, 2, 3, 4, 5, 6, 7

If you take the middle number, 4. You make it so that every number becomes 4. Remove 1 from 5 and give it to 3. Remove 2 from 6 and give it to 2, remove 3 from 7 and give it to 1 to get:

4, 4, 4, 4, 4, 4, 4

The sum is then the 4n = 4(7) = 28

But what did you do actually? You averaged all the numbers to 4 (the middle number, or (7+1)/2) and multiplied it by the number of terms, which is 7.

Does that make it any clearer?

- September 21st 2011, 02:22 AM #3

- September 21st 2011, 02:48 AM #4

- Joined
- Sep 2010
- From
- Oslo
- Posts
- 57
- Thanks
- 4

- September 21st 2011, 02:51 AM #5

- Joined
- Sep 2010
- From
- Oslo
- Posts
- 57
- Thanks
- 4

- September 21st 2011, 02:58 AM #6
## Re: 1+2+3+...+n

Take a look here:

Mathematical induction - Wikipedia, the free encyclopedia

Your exercice is used as an example.

- September 21st 2011, 06:21 AM #7

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 11,914
- Thanks
- 779

## Re: 1+2+3+...+n

Hello, TwoPlusTwo!

Here is a geometric demonstration of the rule (not a proof, mind you).

Consider the case: .

We have this array of objects:

. .

Left-justify the objects:

. .

. .

. .

. .

. .

Append an inverted copy of the array:

. .

We see that the rectangle has: . objects.

Therefore, the triangle has: . objects.

- September 21st 2011, 06:22 AM #8

- Joined
- Apr 2005
- Posts
- 16,449
- Thanks
- 1864

## Re: 1+2+3+...+n

There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.

Here is how he was supposed to have done it: write

1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it

100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1

and add each**column**. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc.**Every**pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the**two**sums together add to 100(101)= 10100. Since that is**two**sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)

If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.

- September 21st 2011, 07:25 AM #9

- Joined
- Apr 2005
- Posts
- 16,449
- Thanks
- 1864