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Chapter 1: Problem 9

Prove that \(n^{3}+(n+1)^{3}+(n+2)^{3}\) is divisible by 9 for all \(n \in \mathbb{N}\).

Short Answer

Expert verified
By mathematical induction, we proved that the formula \(n^{3}+(n+1)^{3}+(n+2)^{3}\) is divisible by 9 for all natural numbers \(n\).

Step by step solution

01

Showing the base case

To prove the base case, put \(n = 0\) into the formula, resulting in \(0^{3} + 1^{3} + 2^{3} = 0 + 1 + 8 = 9\), and since 9 is divisible by 9, the base case is demonstrated.
02

Assume the formula is valid for \(n\)

Now let's assume the formula \(n^{3} + (n+1)^{3} + (n+2)^{3}\) is divisible by 9 for a certain \(n\). This means there is an integer \(k\) such that \(n^{3} + (n+1)^{3} + (n+2)^{3} = 9k\).
03

Show it's valid for \(n+1\)

Next, try to show the formula is also divisible by 9 for \(n+1\). Start with the left hand side, \((n+1)^{3} + (n+2)^{3} + (n+3)^{3}\). We want to express this in terms of the original formula, \(n^{3} + (n+1)^{3} + (n+2)^{3}\). Expanding \((n+1)^{3}\), \((n+2)^{3}\) and \((n+3)^{3}\) and adding them together, we get \(3n^{3} + 9n^2 + 15n + 9\). Now, if we separated the terms like \(n^{3} + (n+1)^{3} + (n+2)^{3} + n^3 + 3n^2 + 3n + 9 = 6n^3 + 12n^2 + 18n + 27\). The left part is divisible by 9 by our assumption, and the right side is divisible by 9 as well. Thus, the formula is valid for \(n+1\).
04

Conclusion

Because the problem meets all the conditions for mathematical induction (The base case holds and if the formula holds for one number, it will hold for the next), we can conclude that \(n^{3}+(n+1)^{3}+(n+2)^{3}\) is divisible by 9 for all \(n \in \mathbb{N}\).

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