Asteroid sum

We can rewrite the given summation as:

$\lim_{n \to \infty} \dfrac{1}{n} \sum_{r=0}^{n} \left(\dfrac{r}{n}\right)^2 \sqrt{1-\left(\dfrac{r}{n}\right)^3}$

This is a Riemann sum, and can be expressed as the integral:

$I = \int_{0}^{1} x^2 \sqrt{1-x^3} dx$

The value of this integral is $\frac{2}{9}$ and so the required answer is:

$2+9+2 \times 9 = \boxed{29}$

$\begin{aligned} L & = \lim_{n \to \infty} \sum_{r=0}^n \frac {r^2}{n^\frac 92} \sqrt{n^3-r^3} \\ & = \frac 1n \lim_{n \to \infty} \sum_{r=0}^n \left(\frac rn\right)^2 \sqrt{1-\left(\frac rn\right)^3} & \small \color{#3D99F6} \text{It is a Riemann sum} \\ & = \int_0^1 x^2 \sqrt{1-x^3} dx & \small \color{#3D99F6} \lim_{n \to \infty} \frac 1n \sum_{k=a}^b f \left(\frac kn\right) = \lim_{n \to \infty} \int_\frac an^\frac bn f(x) \ dx \\ & = \int_0^\frac \pi 2 \frac 23 \sin \theta \cos^2 \theta \ d\theta & \small \color{#3D99F6} \text{Let }x = \sin^\frac 23 \theta \implies dx = \frac 23 \sin^{-\frac 13} \theta \cos \theta \ d\theta \\ & = \frac 23 \int_0^1 u^2\ du & \small \color{#3D99F6} \text{Let }u = \cos \theta \implies du = - \sin \theta \ d\theta \\ & = \frac 29 \end{aligned}$

Therefore $a+b+ab = 2+9+2\times 9 = \boxed{29}$ .

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Reference: Riemann sum