91Ó°ÊÓ

Let's start by imagining a three-dimensional checkerboard. This isn't like a regular 2D game board—it's more like a Rubik's cube.
For this problem, we consider a large cube divided into smaller unit cubes. Specifically, we deal with a cube of size \(2^n \times 2^n \times 2^n\), where \(n\) is a positive integer.
Now, taking out one of these tiny unit cubes results in a slightly imperfect 3D checkerboard. Our ultimate goal is to cover this imperfect 3D checkerboard using only \(2 \times 2 \times 2\) cubes that also have one unit cube missing.
This is a classic example of a problem in higher dimensions, emphasizing spatial thinking and pattern recognition.

In any induction proof, we start with what is called the 'base case.' It's the simplest version of the problem.
For \(n=1\), our 3D checkerboard becomes \(2^1 \times 2^1 \times 2^1 = 2 \times 2 \times 2\). This is an 8-unit cube.
Remove one unit cube, and you have 7 remaining. We need to check if these 7 remaining cubes can be covered with one \(2 \times 2 \times 2\) cube that also has one unit cube removed.
It's clear that they can be covered, as there is exactly one \(2 \times 2 \times 2\) cube left with one missing unit. This satisfies our base case.

With the base case confirmed, we move to the inductive hypothesis. This step is like saying, 'Assume the rule works for some number \(k\).'
Here, we assume that any \(2^k \times 2^k \times 2^k\) checkerboard with one missing unit cube can be covered by \(2 \times 2 \times 2\) cubes missing one cube each.
This assumption helps us bridge the gap to prove it for the next number, \(k+1\).

Next is the inductive step, which essentially says, 'If our assumption works for \(k\), it should work for \(k+1\) as well.'
Consider the larger \(2^{k+1} \times 2^{k+1} \times 2^{k+1}\) cube. We can break this larger cube into 8 smaller \(2^k \times 2^k \times 2^k\) cubes.
By our inductive hypothesis, each of these smaller cubes, with one cube missing, can be covered by \(2 \times 2 \times 2\) cubes missing one unit each.
Therefore, the entire larger cube with only one missing unit can be covered.
This completes the inductive step, ensuring that our covering rule works for any \(n\).
In conclusion, thanks to mathematical induction, we confirm that a \(2^n \times 2^n \times 2^n\) checkerboard with one cube missing can indeed be covered by \(2 \times 2 \times 2\) cubes with one cube removed, for any \(n \geq 1\).