If the product $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9$, what is the sum of $a$ and $b$?
$\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37$
Notice that the numerator of the first fraction cancels out the denominator of the second fraction, and the numerator of the second fraction cancels out the denominator of the third fraction, and so on.
The only numbers left will be $a$ in the numerator from the last fraction and $2$ in the denominator from the first fraction.  (The $b$ will cancel with the numerator of the preceeding number.)  Thus, $\frac{a}{2} = 9$, and $a=18$.
Since the numerator is always one more than the denominator, $b = a-1 = 17$, and $a+ b = 18 + 17 = 35$, giving an answer of $\boxed{D}$