<s>[INST] Proof by induction that $4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}$ I am stuck on this problem for my discrete math class.

Prove the equation by induction for all integers greater than or equal to $3$:
$$4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}.$$

I know that base case $n=3$:
$4^3=64$ as well as $4(4^3-16)/3 = 64$
My confusion is on the induction step where:
$4^3+4^4+4^5+⋅⋅⋅+4^n+4^{(n+1)} = 4(4^{(n+1)}-16)/3$.
I don't know what to do next.
 [/INST] By the induction hypothesis 
$$
4^3+4^4+\ldots+4^n=\frac{4(4^n-16)}{3}=\frac{4^{n+1}-64}{3}
$$
so 
$$
4^3+4^4+\ldots+4^n+4^{n+1}=\frac{4^{n+1}-64}{3}+4^{n+1}=\frac{4^{n+1}-64+3\cdot4^{n+1}}{3}\\
=\frac{4\cdot 4^{n+1}-64}{3}=\frac{4^{n+2}-64}{3}=\frac{4(4^{n+1}-16)}{3}
$$
QED
</s>