Without using Binomial Theorem

Given that $\displaystyle (1+2x)^n = \sum_{i=0}^n a_i x^i$ and $\displaystyle (1+2)^n = \sum_{i=0}^n a_i = 243 = 3^5 \implies n = 5$ . Now we have:

$\begin{aligned} \sum_{i=0}^5 a_i x^i & = (1+2x)^5 & \small \blue{\text{Integrate both sides w.r.t. }x} \\ \sum_{i=0}^5 \frac {a_i x^{i+1}}{i+1} & = \int (1+2x)^5 dx & \small \blue{\text{Putting }x=1} \\ \sum_{i=0}^5 \frac {a_i}{i+1} & = \int_0^1 (1+2x)^5 dx & \small \blue{\text{Let }u = 1+2x} \\ & = \frac 12 \int_1^3 u^5 du & \small \blue{\implies du = 2\ dx} \\ & = \frac 12 \left[ \frac {u^6}6 \right]_1^3 = \frac {3^6-1}{12} = \frac {182}3 \end{aligned}$

Therefore $p+q = 182+3 = \boxed{185}$ .

Substituting $x=1$ yields $\displaystyle 3^n=\sum_{1=0}^n a_{i}=243$ which implies $n=5$ .

Hence integrating on both sides from $0$ to $1$ yields $\displaystyle \int_{0}^{1}(1+2x)^{5} dx=\sum_{i=0}^5 \dfrac{a_i}{i+1}=\boxed{\dfrac{182}{3}}$ hence making the answer $\boxed{185}$ .