Question

What is the sum of n natural numbers: 1+2+...+n?

Answer 1

If

$S=1+2+\cdots+(n-1)+n$

then we also have

$S=n+(n-1)+\cdots+2+1$

Pairing up the terms, we can find that

$2S=(1+n)+(2+n-1)+\cdots+(n-1+2)+(n+1)$

$2S=(n+1)+(n+1)+\cdots+(n+1)+(n+1)$

$S$ consists of $n$ terms, so there are $n$ copies of $n+1$ in the sum $2S$. Then we find an immediate closed form for the sum:

$2S=n(n+1)\implies S=\dfrac{n(n+1)}2$