- First part of the task is just to show that $(4^n-1)$ actually is divisible by 3 for n=1,2,3,4. No problem.
-
Second step: is to show that $(4^n -1) = (2^n-1)(2^n+1)$ No problem, just algebra.
-
Third step is to explain that $(2^n-1)$,$2^n$,$(2^n+1)$ is three consecutive numbers. And that only one is divisible by three.
$2^n$ has two as a factor and is not divisible by 3. It can't be even. That leaves $(2^n-1)$ and $(2^n+1)$. The one with 3 as a factor seems to be random dependent on n. -
Fourth step, tie it all together.
Now, I'm not sure if I'm suppose to dig deeper into which of them is actually divisible by 3. Or has 3 as a factor. But knowing that one of them at the time (dependent on n) has indeed 3 as a factor implies that, in respect to the second step, 3 is a factor of $(4^n -1)$.
Is this all there is to it? Keep in mind that this is a really beginners proof task, not very formal. It's slightly funny going back to high school math after being scared to death by logic and descrete math at university-level. I just expect everything thing to be super complex.
Best Answer
$(4^n-1)=(4-1)(4^{n-1}+4^{n-2}+...+4^{0})$