Voltage increase per cycle across capacitor for a boost converter
Energy stored in a capacitor is \$ \frac{1}{2}CV^2 \$.

Energy stored in a inductor is \$ \frac{1}{2}LI^2 \$.

So the conservation of energy equation would be:
$$ \frac{1}{2}LI_i^2 - \frac{1}{2}LI_f^2 = \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2 $$
(subscript i means initial, f means final)

Using an assumption that \$V_{out}\$ is near constant over one cycle, that is \$ V_{out} \gg V_f-V_i \$, then \$ 2 V_{out} \approx V_{f} + V_{i} \$.
$$ \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2 = \frac{1}{2}C(V_f^2-V_i^2) 
= \frac{1}{2}C(V_f+V_i)(V_f-V_i) = C V_{out} \Delta V$$
This is the answer given by jp314.