Observe the following pattern
$$2^2-1^2=2+1$$
$$3^2-2^2=3+2$$
$$4^2-3^2=4+3$$
$$5^2-4^2=5+4$$
and find the value of
$$100^2-99^2$$
Correct option is A. $$199$$
From the pattern, we can say that the difference between the squares of two consecutive numbers is sum of the numbers itself.
$$(n+1)^2-n^2=(n+1)+n$$
Using above formula we get,
$$100^2-99^2=(99+1)+99$$
$$=100+99$$
$$=199$$