Answer
$S_n=\dfrac{n(1+n)}{2}$
Work Step by Step
He could reason as follows.
Let's rewrite the sum $1+2+3+ . . . +100$ backward
$S=1+2+3+4 + . . . + 100$
$S=100 + 99 + 98+97 + . . . +1$
Let's add the two sums by adding the terms in a pair
$2S=(1+100) + (2+99) +(98+3) +(97+4)+ . . .+(100+1)$
There are 100 such pairs in total and the sum of each pair is 101.
$2S=100\cdot101$ or $S=\dfrac{100\cdot101}{2}=5050$
If there were $n$ terms, then the pairs would be $n$ as well and the sum in each pair would be $n + 1$. So for the sum $S_n=1+2+3+4+ . . . + n$ we have $S_n=\dfrac{n(1+n)}{2}$
