In my text book,it asks us to prove the statement below using induction. My question here isn't about answering that, but rather about how they got the sequence.
The problem shows: For integers $$n \geq 3$$ $$\displaystyle 4^3+4^4+4^5+\cdot \cdot \cdot + 4^n = \frac{4(4^n -16)}{3}$$
Now we know that for a geometric progression, the formula for $$ar^k$$ looks somethings like: $$\displaystyle \sum_{k =0}^{n} ar^k = \frac{ar^{n+1} -a}{r-1} , r \neq 1$$
Now how did did exactly did we arrive at this? If we plug 4 in as a and as r we get $$\displaystyle \frac{4(4^{n+1} -4)}{3}$$ Which is close, but has a -4 instead of a -16 in the numerator.
The series $\sum_{k=0}^n ar^k$ is $a+ar+ar^2+\ldots+ar^n$ so the first term is $a$, and there are $n+1$ terms in total. Note that $$4^3+4^4+\ldots+4^n = 4^3(1+4+4^2+\ldots+4^{n-3}).$$ Using the formula, $$1+4+4^2+\ldots+4^{n-3}=\sum_{k=0}^{n-3}4^k=\frac{4^{n-2}-1}{4-1}=\frac{4^{n-2}-1}{3}$$ Therefore $$4^3+4^4+\ldots+4^n =4^3\times\frac{4^{n-2}-1}{3}= \frac{4^{n+1}-4^3}{3}=\frac{4(4^n-16)}{3}$$