Can this inductive proof that $\sum_{i=0}^n2^{2i+1}=\frac23(4^n-1)$ be simplified? The general structure of equations I've used for the inductive step for proofs with a summation is something like:
We'll prove that $\sum_{i = 0}^{n + 1} (\text{something}) = (\text{closed form expression})$
\begin{align}
\sum_{i = 0}^{n + 1} (\text{something}) &= \sum_{i = 0}^n (\text{something}) + \text{last term} &\\
&= [\text{expression via I.H.}] + \text{last term} &\\
&= \text{do some work...} &\\
&= \text{some more work...} &\\
&= (\text{finally reach the closed form expression we want})
\end{align}
This structure is very nice, since the equation is one-sided, and very easy to follow. However I solved a problem that I couldn't solve with this one-sided structure, and I had to substitute the LHS with the closed form expression I'm trying to prove, so I could use some of its terms to simplify the RHS. This is fine and valid, but I'd like to know if there's a simpler way to perform this proof that doesn't employ the substitution you see below:

In other words, I couldn't figure out how to simplify $\frac{2}{3}(4^n - 1) + 2^{2n + 1}$ to get $\frac{2}{3}(4^{n + 1} - 1)$. The farthest I got was:
\begin{align}
&= \frac{2}{3}(4^n - 1) + 2^{2n + 1} &\\
&= \frac{2}{3}(4^n - 1) + 2\cdot 2^{2n} &\\
&= \frac{2}{3}(4^n - 1) + 2\cdot 4^n &\\
&= \frac{2}{3}(4^n - 1) + \frac{3 \cdot 2\cdot 4^n}{3} &\\
&= \frac{2}{3}(4^n - 1 + 3 \cdot 4^n) &\\
\end{align}

You want to prove
$$\frac23(4^{n+1}-1)-\frac23(4^n-1)=2^{2n+1}.$$
Simplifying the $-1$ and dividing by $2\cdot4^n$,
$$\frac13\cdot4-\frac13=1.$$

Note that this is just as case of the sum of a geometric progression
$$\sum_{k=0}^{n-1}ab^k=a\frac{b^n-1}{b-1},$$ proven by
$$ab^n=a\frac{b^{n+1}-1}{b-1}-a\frac{b^n-1}{b-1}=ab^n\frac{b-1}{b-1}.$$
