3. Among the numbers $29998$, $29999$, $30000$, $30001$, the one that can be expressed as the sum of the products of three consecutive natural numbers taken two at a time is ( ).
(A) 30001
(B) 30000
(C) 29999
(D) 29998
3. C.

Notice that, for three consecutive natural numbers $n-1, n, n+1$, the sum of their pairwise products has the form
$$
n(n-1)+n(n+1)+(n-1)(n+1)=3 n^{2}-1 \text {. }
$$

The above expression leaves a remainder of 2 when divided by 3, while 30001 and 29998 leave a remainder of 1 when divided by 3, and 30000 is divisible by 3. Thus, we can exclude options $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{D}$.
In fact, 29999 leaves a remainder of 2 when divided by 3, and
$$
99 \times 100+100 \times 101+101 \times 99=29999 .
$$