My way before is unnecessarily long. When you add $(1 + 2 + \dots + n) + (n + (n-1) + \dots + 1)$, line up the pairs $(n, 1)$, $(2, n-1)$, $(3, n-2)$, and you'll notice that they all sum to $n+1$. Since you have $n$ pairs, the sum is $n(n+1)$.
My way before is unnecessarily long. When you add $(1 + 2 + \dots + n) + (n + (n-1) + \dots + 1)$, line up the pairs $(n, 1)$, $(2, n-1)$, $(3, n-2)$, and you'll notice that they all sum to $n+1$. Since you have $n$ pairs, the sum is $n(n+1)$.