$4(299)+3(299)+2(299)+298=$
$\text{(A)}\ 2889 \qquad \text{(B)}\ 2989 \qquad \text{(C)}\ 2991 \qquad \text{(D)}\ 2999 \qquad \text{(E)}\ 3009$
We can make use of the [distributive property](https://artofproblemsolving.com/wiki/index.php/Distributive_property) as follows:
\begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \\ &= (4+3+2+1)(299)-1 \\ &= 10(299)-1 \\ &= 2989 \\ \end{align*}
$\boxed{\text{B}}$