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**a proof of why this is the sum of all naturals **

I’ve done the proof just wondering is there a proof that shows more intuitively why this ends up being gauss’s formula.

Bradley Rideout
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    It's actually off by a shift. Sum of first n positive integers is $(n+1)n/2$ so $\binom{n+1}{2}/$ – coffeemath Jan 12 '22 at 20:13
  • The factorial manipulation always distracts people from the purpose. You have $n$ choices of a first object, $n-1$ choices of a second. But then if you pick A and B or B and A you should get the same answer, so you divide by 2 to correct for that double count. I think the problem is people learning $\binom{n}{k}$ in terms of factorials without internalizing why the combinatorial and factorial definitions coincide, leading to messy proofs of things like this. –  Jan 12 '22 at 20:18
  • Does this answer your question? Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ – Hans Lundmark Jan 12 '22 at 21:28

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Here is a combinatorial proof that $$1+2+\cdots + n = \binom{n+1}{2}.$$

Imagine you have $n+1$ objects. Lay them in a line. Pick two as follows: first pick one object, then pick a second object to the right of your initial object.

If you pick the leftmost object as your initial one, you have $n$ ways of picking the second object. If you pick the second object from the left first, you have $n-1$ ways. So on, and so in total you have $$n + (n-1) + \cdots + 1 + 0$$ ways to pick $2$ objects from a total of $n+1.$

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By your result, $\binom{n}{2}=\frac{n(n-1)}{2}$ which is the formula for the sum of the numbers $1,2,...n-1$.

You can see why this is true by writing out the combinations:

$(1,2), (1,3), ..., (1,n)$ is $n-1$ combinations

$(2,3),(2,4),..,(2,n)$ is $n-2$ combinations

$\vdots$

$(n-1, n)$ is $1$ combination.

Add them up to get the result.

John Douma
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