$\dfrac{2+4+6+\cdots + 34}{3+6+9+\cdots+51}=$
$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{2}{3} \qquad \text{(C)}\ \dfrac{3}{2} \qquad \text{(D)}\ \dfrac{17}{3} \qquad \text{(E)}\ \dfrac{34}{3}$
First, notice that each number in the numerator is a multiple of $2$, and each number in the denominator is a multiple of $3$.  This suggests that each expression can be factored.  Factoring gives:
$\dfrac{2(1+2+3+\cdots + 17)}{3(1+2+3+\cdots + 17)}$
Since all the material in the parentheses is the same, the common factor in the numerator and the denominator may be cancelled, leaving $\frac{2}{3}$, which is option $\boxed{B}$.