Prove by mathematical induction that $4^k - 1$ is divisible by $3$.
$p = (4^k-1) \bmod 3 = 0$.
If $k = 1$, $(4^k - 1) \bmod 3 = 0$, which is true.
Assuming $4^k - 1$ is true, lets continue to the induction step:
$(4^k \cdot 4^1) - 1$
$3\cdot 4^k + (4^k - 1)$
$3 \cdot 4^k$ is a multiple of 3, and $4^k - 1$ is true. Which means that the statement $p$ is true.
I understand everything up to the 2nd step where they have $(4^k \cdot 4^1) -1$. My question is why add $4^k-1$ to $3 \cdot 4^k$ in the 3rd step? What is the reasoning behind this?