Understanding the Problem

We are given two sets:

• X = {4ⁿ – 3n – 1 : n ∈ N}
• Y = {9(n – 1) : n ∈ N}

where N is the set of natural numbers {1, 2, 3, ...}. We need to find what X ∪ Y equals.

Step 1: List the Elements of Set Y

Y = {9(n – 1) : n ∈ N}

For n = 1: 9(1 – 1) = 0

For n = 2: 9(2 – 1) = 9

For n = 3: 9(3 – 1) = 18

For n = 4: 9(4 – 1) = 27

So, Y = {0, 9, 18, 27, 36, ...} which is the set of all non-negative multiples of 9.

Step 2: List the First Few Elements of Set X

X = {4ⁿ – 3n – 1 : n ∈ N}

For n = 1: 4¹ – 3(1) – 1 = 4 – 3 – 1 = 0

For n = 2: 4² – 3(2) – 1 = 16 – 6 – 1 = 9

For n = 3: 4³ – 3(3) – 1 = 64 – 9 – 1 = 54

For n = 4: 4⁴ – 3(4) – 1 = 256 – 12 – 1 = 243

For n = 5: 4⁵ – 3(5) – 1 = 1024 – 15 – 1 = 1008

So, the first few elements of X are {0, 9, 54, 243, 1008, ...}

Step 3: Compare Sets X and Y

From the calculated values:

• 0 is in both X and Y.
• 9 is in both X and Y.
• 18 is in Y but not in X (from our calculation, X has 54 next, not 18).
• 27 is in Y but not in X.
• 36 is in Y but not in X.
• 54 is in X but not in Y (Y has multiples of 9: 0,9,18,27,36,45,54,... Wait, 54 is a multiple of 9, so it should be in Y as well. Let's check: For Y to contain 54, we need 9(n-1)=54 ⇒ n-1=6 ⇒ n=7. So yes, 54 ∈ Y).

This suggests that many elements of X are also in Y. Let's investigate this pattern.

Step 4: Prove that X is a Subset of Y

We need to show that for every natural number n, the number 4ⁿ – 3n – 1 is divisible by 9, meaning it is a multiple of 9 and therefore an element of Y.

We can prove this using mathematical induction.

Base Case (n=1): 4¹ – 3(1) – 1 = 0, which is divisible by 9. True.

Inductive Hypothesis: Assume for n = k, that 4^k – 3k – 1 is divisible by 9. So, 4^k – 3k – 1 = 9m, for some integer m.

Inductive Step (n = k+1): We need to show 4^k+1 – 3(k+1) – 1 is divisible by 9.

4^k+1 – 3(k+1) – 1 = 4 * 4^k – 3k – 3 – 1 = 4 * 4^k – 3k – 4

From the inductive hypothesis, 4^k = 9m + 3k + 1. Substitute this in:

= 4(9m + 3k + 1) – 3k – 4

= 36m + 12k + 4 – 3k – 4

= 36m + 9k

= 9(4m + k)

This is clearly divisible by 9. Therefore, by induction, every element of X is a multiple of 9, meaning X ⊆ Y.

Step 5: Find the Union X ∪ Y

The union of two sets is the set of all elements that are in X, or in Y, or in both.

Since we have proven that X is a subset of Y (X ⊆ Y), this means every element of X is already contained in Y.

Therefore, the union X ∪ Y is simply equal to the larger set Y.

X ∪ Y = Y

Final Answer

X ∪ Y is equal to Y.

Related Concepts & Formulas

Set Theory:

• Union (A ∪ B): The set of elements that are in A, or in B, or in both.
• Subset (A ⊆ B): If every element of set A is also an element of set B.
• Property: If A ⊆ B, then A ∪ B = B.

Mathematical Induction:

• A method to prove a statement P(n) is true for all natural numbers n.
• Base Step: Prove P(1) is true.
• Inductive Step: Assume P(k) is true (Inductive Hypothesis), then prove that P(k+1) is true.

Divisibility:

• A number $a$ is divisible by 9 if $a=9k$ for some integer $k$.