2 Answers

Fennie · Answer 1 · 2020-06-27T15:29:22+0000

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

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

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

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

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

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