# 1+2+3+...+n

• Sep 21st 2011, 01:14 AM
TwoPlusTwo
1+2+3+...+n
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.
• Sep 21st 2011, 01:22 AM
Unknown008
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:

$\displaystyle S_n = \dfrac{n}{2}\left(a+l\right)$

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

$\displaystyle S_n = \dfrac{n(n+1)}{2}$

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?
• Sep 21st 2011, 01:22 AM
Siron
Re: 1+2+3+...+n
Do you know 'induction'?
• Sep 21st 2011, 01:48 AM
TwoPlusTwo
Re: 1+2+3+...+n
Yes, that was very clear. Thank you!
• Sep 21st 2011, 01:51 AM
TwoPlusTwo
Re: 1+2+3+...+n
Quote:

Originally Posted by Siron
Do you know 'induction'?

No, I don't. Does that relate to this problem?
• Sep 21st 2011, 01:58 AM
Siron
Re: 1+2+3+...+n
Quote:

Originally Posted by TwoPlusTwo
No, I don't. Does that relate to this problem?

Take a look here:
Mathematical induction - Wikipedia, the free encyclopedia
Your exercice is used as an example.
• Sep 21st 2011, 05:21 AM
Soroban
Re: 1+2+3+...+n
Hello, TwoPlusTwo!

Quote:

$\displaystyle \text{Can anyone tell me why: }\:1+2+3+\hdots +n\:=\:\frac{n(n+1)}{2}$

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

Consider the case: .$\displaystyle n = 5$

We have this array of objects:

. . $\displaystyle \begin{array}{c}\circ \\ \circ\;\circ \\ \circ\;\circ\;\circ \\ \circ\;\circ\;\circ\;\:\circ \\ \circ\;\circ\;\circ\;\circ\;\circ \end{array}$

Left-justify the objects:

. . $\displaystyle \circ$
. . $\displaystyle \circ\;\:\circ$
. . $\displaystyle \circ\;\circ\;\circ$
. . $\displaystyle \circ\;\circ\;\circ\;\:\circ$
. . $\displaystyle \circ\;\circ\;\circ\;\circ\;\circ$

Append an inverted copy of the array:

. . $\displaystyle \begin{array}{c}\circ\;\bullet\;\bullet\;\bullet\; \bullet \;\:\bullet \\ \circ\;\circ\;\bullet\;\bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\; \bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\;\circ\;\bullet\;\:\bullet \\ \circ\;\circ\; \circ\;\circ\;\circ\;\:\bullet \end{array}$

We see that the rectangle has: .$\displaystyle 5 \times 6\:=\:30$ objects.

Therefore, the triangle has: .$\displaystyle \frac{5 \times 6}{2} \:=\:15$ objects.

• Sep 21st 2011, 05:22 AM
HallsofIvy
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.
• Sep 21st 2011, 06:25 AM
HallsofIvy
Re: 1+2+3+...+n
Quote:

Originally Posted by HallsofIvy
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.

As usual, Soroban got in just before me. I need to learn to type faster!