Answer
See below.
Work Step by Step
Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here:
1) For $n=1: 1=\frac{1}{3}(4^1-1)$.
2) Assume for $n=k: 1+4+...+4^{k-1}=\frac{1}{3}(4^k-1)$. Then for $n=k+1$:
$1+4+...+4^{k-1}+4^k=\frac{1}{3}(4^k-1)+4^k=\frac{4}{3}\cdot4^k-\frac{1}{3}=\frac{1}{3}(4^{k+1}-1).$
Thus we proved what we wanted to.