Sum

The sequence of numbers forms an arithmetic progression with a common difference of $d=99$ .

The number of terms is $n=\dfrac{20000}{99}=202$ . The last term is $a_{202}=99(202)=19998$ . The first term is $a_1=99$ . So the terms are $(99,198,297...19998)$ .

By using the formula for the sum, we get

$s=\dfrac{n}{2} \left(a_1+a_{202} \right)=\dfrac{202}{2}(99+19998)=2029797$