Hello Jay,
We are given to prove by induction:
$$\sum_{j=0}^n\left(4^j \right)=\frac{4^{n+1}-1}{3}$$
First, let's verify the base case $P_1$ is true:
$$\sum_{j=0}^1\left(4^j \right)=\frac{4^{1+1}-1}{3}$$
$$4^0+4^1=\frac{4^2-1}{3}$$
$$1+4=\frac{15}{3}$$
$$5=5$$
The base case is true, so let's state the induction hypothesis $P_k$:
$$\sum_{j=0}^k\left(4^j \right)=\frac{4^{k+1}-1}{3}$$
As our inductive step, let's add $$4^{k+1}$$ to both sides:
$$\sum_{j=0}^k\left(4^j \right)+4^{k+1}=\frac{4^{k+1}-1}{3}+4^{k+1}$$
$$\sum_{j=0}^{k+1}\left(4^j \right)=\frac{4^{k+1}-1+3\cdot4^{k+1}}{3}$$
$$\sum_{j=0}^{k+1}\left(4^j \right)=\frac{4\cdot4^{k+1}-1}{3}$$
$$\sum_{j=0}^{k+1}\left(4^j \right)=\frac{4^{(k+1)+1}-1}{3}$$
We have derived $P_{k+1}$ from $P_k$, thereby completing the proof by induction.