Answer
\[
\boxed{P_n = P_0 \,\biggl(1 + \frac{i}{m}\biggr)^{n}.}
\]
Work Step by Step
Given the recurrence
\[
P_k = \left(1 + \frac{i}{m}\right) P_{k-1},
\quad P_0 \text{ is the initial deposit},
\]
we see this is a geometric progression with common ratio \(r = 1 + \tfrac{i}{m}\). Thus, by iterating:
\[
P_k
= \left(1 + \frac{i}{m}\right) P_{k-1}
= \left(1 + \frac{i}{m}\right)\bigl[\left(1 + \frac{i}{m}\right) P_{k-2}\bigr]
= \dots
= \left(1 + \frac{i}{m}\right)^k P_{0}.
\]
