Ambiguity of Small and Large

$5x^5=x^4$ or, $x^4(x-\frac{1}{5})=0$ . So, $r=\frac{1}{5}$ .

Now, $\frac{p}{q}=3r^2+4r^3+5r^4+6r^5+\ldots \\ =(1+2r+3r^2+4r^3+5r^4+6r^5+\ldots)-(1+2r) \\ =(1-r)^{-2}-(1+2r) =\frac{25}{16}-(1+\frac{2}{5}) =\frac{13}{80}$ .

So, $p+q=13+80=\boxed{93}$

$f(x) = 5x^5-x^4$

Using Vieta's, sum of roots = $\dfrac 15$ . So $r = \dfrac 15$

Given sum is $\begin{aligned}S &= 3x^2 + 4x^3 + 5x^4 + \cdots\\ &= \sum_{k=3}^\infty kx^{k-1}\\&=\dfrac d{dx} \sum_{k=3}^\infty x^k\\&=\dfrac d{dx} \left\{\dfrac{x^3}{1-x}\right\}\\&=\dfrac{3(1-x)x^2+x^3}{(1-x)^2}\end{aligned}$

So, $\dfrac p q = \dfrac{3\cdot 4 + 1}{5\cdot 16}=\dfrac {13}{80}$

Hence, $\color{#3D99F6}{p+q = \boxed{93}}$

Relevant wiki: Arithmetic and Geometric Progressions Problem Solving

$5x^5 = x^4 \quad \implies \quad x \in \left\{ 0 , \frac{1}{5} \right\}$ $\therefore \boxed{ r = \frac{1}{5} }$

Let, $S = \frac{p}{q} = 3r^2 + 4r^3 + 5r^4 + 6r^5 + \cdots$ $\therefore r\times S = 3r^3 + 4r^4 + 5r^5 + 6r^6 + \cdots$ $\therefore (1 - r) \times S = 3r^2 + r^3 + r^4 + r^5 + \cdots$

Remember Geometric Progression

$\implies S = \frac{1}{1 - r} \times \left( 3r^2 + r^3\frac{1}{1 - r} \right) = \frac{r^2(3-2r)}{(1 - r)^2}$ $\therefore S = \frac{13}{80} = \frac{p}{q}$ $\boxed{p+q = 93}$

$r = \frac{1}{5} \\\ \text {by vieta's formula.}$

let $S = 3r^{2} + 4r^{3} + ...$

$\rightarrow rs = 3r^{3} + 4r^{4} + ...$

$\rightarrow S(1-r) = 3r^{2} + r^{3} + r^{4} + ...$

$\rightarrow S(1-r) = 3r^{2} + \dfrac{r^{3}}{1-r}$

substituting $r = \frac{1}{5}$

$\rightarrow S(\frac{4}{5}) = 3(\frac{1}{5})^{2} + \dfrac{\frac{1}{125}}{\frac{4}{5}}$

$\rightarrow S(\frac{4}{5}) = \frac{3}{25} + \frac{1}{100}$

$\rightarrow S = \frac{3}{20} + \frac{1}{80} = \frac{13}{80}$

$\rightarrow \boxed{p + q = 13 + 80 = 93}$ .