Algebra Question

Given equation is

$\underset{i=0}{\overset{n}{\sum 4^i}}=\frac{1}{3}\left(4^{n+1}-1\right), n\ge 0$

For n=0

$$\begin{gathered} \underset{i=0}{\overset{0}{\sum 4^i}}=\frac{1}{3}\left(4^{0+1}-1\right) \\ {4}^0=\frac{1}{3}\left(4^1-1\right) \\ 1=\frac{1}{3}\left(4-1\right)=1 \end{gathered}$$

The result is true for n=0.

Let us assume that the equation holds for n=m, that is

$\underset{i=0}{\overset{m}{\sum 4^i}}=\frac{1}{3}\left(4^{m+1}-1\right)$