What is the Fastest Way?

The general "formula" for the sum of the first "n" positive integers is given by:

n(n + 1) / 2   ...so we have.....for the first 10 integers....

(10)(11)/2   = 110/2  = 55

1+2+3+4+5+6+7+8+9+10 = 55  What is the Fastest way to do that?

$$\small{\text{$
1+2+3+4+5+6+7+8+9+10 =
\dbinom{n+1}{2} = \dbinom{10+1}{2} = \dbinom{11}{2}
= \dbinom{11}{9}=\dfrac{10\cdot 11}{2} = 5\cdot 11 = 55
$}}$$