Can someone please explain why 1+2+3+...+n equals (n+1)_C_2 ?

Steven G

Elite Member
Joined
Dec 30, 2014
Messages
14,364
Can someone please explain why 1+2+3+...+n = (n+1)C2? Of course I know that they both equal n(n+1)/2. What I am really asking is why is choosing 2 objects from (n+1) objects the same as adding up the 1st n positive integers?

Being able to prove that two are equal and seeing why the two are equal are not always the same!
 
Can someone please explain why 1+2+3+...+n = (n+1)C2? Of course I know that they both equal n(n+1)/2. What I am really asking is why is choosing 2 objects from (n+1) objects the same as adding up the 1st n positive integers?

Being able to prove that two are equal and seeing why the two are equal are not always the same!

There isn't always a direct answer to a "why" question. But here, at least we can see that both expressions answer the same question.

That question is, given n+1 items, how many pairs are there? As it is commonly asked, if there are n+1 people in a room, and everyone shakes hands with everyone else exactly once, how many handshakes are done? Or, if you put n+1 dots equally spaced around a circle and connect each one to all the others, how many lines do you draw?

Obviously (n+1)C2 answers the question.

But now let's count the pairs. The first person pairs with the n other people. The second person pairs with the first (that pair is already counted) and also with the n-1 people remaining. Continue like that, and the next to last person pairs with the last person; the last person's shakes are already counted.

So the total is n + (n-1) + (n-2) + ... + 2 + 1, which is your sum.

This could be made into a nice colored animation, and probably someone has done it. I didn't invent this!
 
Can someone please explain why 1+2+3+...+n = (n+1)C2? Of course I know that they both equal n(n+1)/2. What I am really asking is why is choosing 2 objects from (n+1) objects the same as adding up the 1st n positive integers?

Being able to prove that two are equal and seeing why the two are equal are not always the same!


Choose 2 out of 4 (A, B, C, D)

A and 1 of 3 (B, C, D) = 3
B and 1 of 2 (C, D) = 2
C and 1 of 1 (D) = 1
1+2+3 = 6 = sum of first 3 ints.
 
Or: \(\displaystyle _{n+1}C_2= \dfrac{(n+1)!}{2(n+1- 2)!}= \dfrac{(n+1)!}{2(n- 1)!}\)

But \(\displaystyle \dfrac{(n+1)!}{(n-1)!}= \dfrac{(n+ 1)(n)(n-1)(n-2)\cdot\cdot\cdot(3)(2)(1)}{(n-1)(n-2)\cdot\cdot\cdot(3)(2)(1)}= (n+1)n\) because all the other terms cancel.
 
Last edited by a moderator:
Can someone please explain why 1+2+3+...+n = (n+1)C2? Of course I know that they both equal n(n+1)/2. What I am really asking is why is choosing 2 objects from (n+1) objects the same as adding up the 1st n positive integers?

Being able to prove that two are equal and seeing why the two are equal are not always the same!
You can check it out with the visual

Code:
1
1 1
1 2  1
1 3  3  1
1 4  6  4  1
1 5 10 10  5  1
1 6 15 20 15  6 1
1 7 21 35 35 21 7 1

Wander on down column 2. When you want to stop, skip to column 3. One might call this a natural development of binomial coefficients, I suppose.
 
Top