Let $I_2=\int_0^1\left(1-x^{50}\right)^{101} d x$, using integration by parts
$$
\begin{aligned}
& \left.=\left[(1-x)^{50}\right)^{101} \cdot x\right]_0^1+\int_0^1\left(1-x^{50}\right)^{100} 50 \cdot x^{49} \cdot x d x \\
& =0-\int_0^1(50)(101)\left(1-x^{50}\right)^{100}(-x)^{50} d x \\
& =-(50)(101) \int_0^1\left(1-x^{50}\right)^{101} d x+(50)(101) \int_0^1\left(1-x^{50}\right)^{100} d x \\
& =5050 I_2+5050 I_1
\end{aligned}
$$

$$
\begin{aligned}
\therefore & I_2+5050 I_2 & =5050 I_1 \\
\therefore & \frac{(5050) I_1}{I_2} & =5051
\end{aligned}
$$