How to solve this power series problem

$$\prod_{n=1}^{100} \left[\frac{x^3}{4^n}+\left(\frac{\pi}{2^n} \right)^2 \right]= \sum_{k=0}^{\infty} a_kx^k$$ Find the coefficients $a_k$ in front of the individual $x^k$ terms for all $k \in \mathbb{N}$

The first thing i tried was using the fact: $$\sum_{k=0}^{\infty} a_kx^k=\frac{1}{1-(a_kx^k)}$$ But failed pretty much after that I evaluated the LHS as $$\frac{(x^3+ \pi^2)^{100}}{4^{5050}}$$

I also tried assuming $a_2=(a_1)^2$ to make it a geometric series, but that didnt fit well with me so i stopped there

well you are nearly done there, just expand the LHS you obtained: $LHS = \frac{(x^3 + \pi^2)^{100}}{4^{5050}} = \frac{\sum_{i = 0}^{100} {100 \choose i} x^{3i} \pi^{2(100 - i)}}{4^{5050}}$ giving $a_{3*k} = \frac{100 \choose k}{4^{5050}} \pi^{200 - 2k}$ where $k \in [0, 100]$ and zero everywhere else