<s>[INST] Prove for all integers n such that n ≥ 3, $ 4^3 + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$ I am trying to prove this using mathematical induction, but I'm lost once I get to comparing the two sides of the equation.
Proposition:
For all integers n such that n ≥ 3,   
$ 4^3  + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$
Proof: 
Let the property P(n) be the equation
$P(n) = 4^3  + 4^4 + 4^5 … 4^n = \frac{4(4^n - 16)}{3}$
Show that P(3) is true:
$4^3 = \frac{4(4^3 - 16)}{3}$
64 = 64, thus P(3) is true
Show that for all integers where n ≥ 3, if P(k) is true, then P(k + 1) is also true:
Suppose that P(k) is true for some particular but arbitrary integer where k ≥ 3.  Suppose that k is any integer where k ≥ 3 such that:
$4^3  + 4^4 + 4^5 … 4^k = \frac{4(4^k - 16)}{3}$
We must show that P(k + 1) is true.  That is, we must show that:
$4^3  + 4^4 + 4^5 … 4^{k + 1} = \frac{4(4^{k + 1} - 16)}{3}$
The left hand side is:
$4^3  + 4^4 + 4^5 … 4^{k + 1}$
$4^3  + 4^4 + 4^5 … 4k + 4^{k + 1}$
$4^3  + 4^4 + 4^5 … \frac{4(4^k - 16)}{3} + 4^{k + 1}$
 [/INST] Using the induction hypothesis, the last line you wrote should be
$\frac{4(4^k - 16)}{3} + 4^{k + 1}$. Then:
\begin{align*}
\frac{4(4^k - 16)}{3} + 4^{k + 1} &= \frac{(4^{k+1} - 4\cdot16)}{3} + \frac {3\cdot4^{k + 1}}3
\\ &= \frac{4^{k+1} - 4\cdot16+3\cdot4^{k + 1}}3
\\ &= \frac{4\cdot 4^{k+1} - 4\cdot16}3
\\ &= \frac{4\cdot (4^{k+1} - 16)}3 \end{align*}
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