So here comes the easy one!

We know that sum of (n) natural numbers is $\dfrac{n(n+1)}{2}$

so , $\dfrac{100(101)}{2}$ = $\boxed{5050}$

$1 + 2 + 3 + ... + 100$

$= (1 + 100) + (2 + 99) + (3 + 98) + ... + (50 + 51)$

$= 101 \times 50$

$\boxed {5050}$

The sum of an Arithmetic Progression is

$S = \frac{n}{2}(a_{1} + a_{n}) = \frac{100}{2}(1+100) = 5050$

The answer is $\dfrac{100 × 101}{2} = 50 × 101 = \color{#69047E}{\boxed{5050}}$ .

$\sum _{ n=1 }^{ 100 }{ n } = 5050$

$1+2+4+\dotsm+n=\frac{n(n+1)}{2}$ Setting $n=100$ , this becomes: $1+2+3+4+\dotsm+100=\frac{100(100+1)}{2}=\frac{100(101)}{2}=50(101)=\boxed{5050}$

Sum of n terms formula of a.p =>Sn=n/2×(2a+(n-1)×d) where a=1, n=100, d=1, put all the values and get the answer