Division Throughout

Rewrite the limit/sum as

$\displaystyle\lim_{n \rightarrow \infty} \left( \dfrac{1}{n} \sum_{r=1}^{n} \dfrac{1}{\sqrt{\frac{r}{n}} (4 + 3\sqrt{\frac{r}{n}})^{2}} \right).$

We can then see that this is a right-hand Riemann approximation of the integral

$\displaystyle\int_{0}^{1} \dfrac{1}{\sqrt{x}(4 + 3\sqrt{x})^{2}} dx.$

As the integrand is Riemann integrable, our limit will be equal to this definite integral.

To solve this integral, let $u = 4 + 3\sqrt{x}.$ Then $du = \dfrac{3}{2\sqrt{x}} dx \Longrightarrow \dfrac{dx}{\sqrt{x}} = \dfrac{2}{3} du.$ Also, as $x$ goes from $0$ to $1$ we have $u$ going from $4$ to $7.$ The integral then becomes

$\dfrac{2}{3} \displaystyle\int_{4}^{7} \dfrac{1}{u^{2}} du = \dfrac{2}{3} \left( -\dfrac{1}{u} \right)_{4}^{7} = \dfrac{2}{3} \left( -\dfrac{1}{7} + \dfrac{1}{4} \right) = \dfrac{2}{3} \times \dfrac{3}{28} = \dfrac{1}{14}.$

Thus $a + b = 1 + 14 = \boxed{15}.$