If X = {4ⁿ - 3n - 1 : n ∈ N} and Y = {9(n-1) : n ∈ N}, where N is the set of natural numbers, then X

Q: Wait, why do we test values first? Isn't that just guessing?

It is not guessing; it is building intuition! In competitive exams like JEE, pattern recognition is a powerful tool. By calculating the first few terms, you form a hypothesis. The subsequent algebraic proof using the Binomial Theorem then confirms that your observation holds true for all $n \in \mathbb{N}$.

Q: Can I use mathematical induction instead of the Binomial Theorem?

Absolutely! Mathematical induction is a perfectly valid and rigorous way to prove that $4^n - 3n - 1$ is divisible by $9$. However, the Binomial Theorem is often faster and more elegant for this specific type of expression, as it reveals the structure of the divisibility directly.

Q: Why is $X \cup Y = Y$ if $X \subseteq Y$?

Think of it in terms of a Venn diagram. If $X$ is a subset of $Y$, then the entire circle of $X$ is contained within the circle of $Y$. When you take the union (everything in $X$ OR $Y$), you are just looking at the entire area covered by $Y$. Thus, the union is simply $Y$.

Question

If $X = \{4^n - 3n - 1 : n \in N\}$ and $Y = \{9(n-1) : n \in N\}$, where N is the set of natural numbers, then $X \cup Y$ is equal to:

Options:

(A) X

(B) Y

(C) N

(D) Y - X

Sigma Percentile · Accepted Answer

### The Beauty of Pattern Recognition

Welcome, future engineer! Today, we are diving into a problem that looks like a simple set theory question but is actually a beautiful dance between algebra and logic.

We have two sets, $X = \{4^n - 3n - 1 : n \in \mathbb{N}\}$ and $Y = \{9(n-1) : n \in \mathbb{N}\}$. At first glance, they seem like strangers. But as we peel back the layers, we find a deep, elegant connection.

### Phase 1: The Detective Work

In any JEE problem, if you are stuck, start by exploring. Let's look at set $Y$ first. It is defined as $9(n-1)$ for natural numbers $n$.

If we plug in $n=1, 2, 3, \dots$, we get $0, 9, 18, 27, \dots$. This is clearly the set of all non-negative multiples of $9$.

Now, let's look at set $X$, defined by $4^n - 3n - 1$. Let's test some values:

- For $n=1$: $4^1 - 3(1) - 1 = 0$
- For $n=2$: $4^2 - 3(2) - 1 = 16 - 6 - 1 = 9$
- For $n=3$: $4^3 - 3(3) - 1 = 64 - 9 - 1 = 54$

Look at these results: $0, 9, 54, 243, \dots$. Every single one of these is a multiple of $9$.

This is our 'Aha!' moment. Our hypothesis is that every element of $X$ is also an element of $Y$, meaning $X \subseteq Y$.

### Phase 2: The Binomial Bridge

To prove this rigorously, we need to show that $4^n - 3n - 1$ is always a multiple of $9$. The expression $4^n$ is the key. We can rewrite $4$ as $(1+3)$, so $4^n = (1+3)^n$.

Now, we invoke the Binomial Theorem:

$$ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots + \binom{n}{n}x^n $$

Substituting $x=3$, we get:

$$ (1+3)^n = \binom{n}{0} + \binom{n}{1}(3) + \binom{n}{2}(3^2) + \binom{n}{3}(3^3) + \dots $$

Since $\binom{n}{0} = 1$ and $\binom{n}{1} = n$, this simplifies to:

$$ (1+3)^n = 1 + 3n + \binom{n}{2}(9) + \binom{n}{3}(27) + \dots $$

### Phase 3: The Elegant Cancellation

Now, let's bring this back to our original expression for $X$: $4^n - 3n - 1$. Substituting our expansion, we get:

$$ X = \left[ 1 + 3n + \binom{n}{2}(9) + \binom{n}{3}(27) + \dots \right] - 3n - 1 $$

Watch the magic happen! The $1$ and $-1$ cancel out. The $3n$ and $-3n$ cancel out. We are left with:

$$ X = \binom{n}{2}(9) + \binom{n}{3}(27) + \dots $$

Every term here contains at least a factor of $9$. We can factor it out:

$$ X = 9 \cdot \left[ \binom{n}{2} + \binom{n}{3}(3) + \dots \right] $$

Since the term in the bracket is an integer, $X$ is indeed a multiple of $9$. This confirms that $X \subseteq Y$.

In the world of sets, if $X$ is a subset of $Y$, then their union $X \cup Y$ is simply the larger set, $Y$. We have solved the puzzle! The final result is **$X \cup Y = Y$**.

Animated Solution for Mathematics - Sets and Relations: If $X = {4^{n} - 3 n - 1 : n \in N}$ and $Y = {9 (n - 1) : n \in N}$ , where N is the set of natural numbers, then $X \cup Y$ is equal to:

Select Answer:

Visualized Solution

The Problem Statement

Analyzing Set $Y$

Elements of Set $Y$

Analyzing Set $X$

Elements of Set $X$

Hypothesis: $X \subseteq Y$

Proving with Binomial Theorem

Expanding $(1 + 3)^{n}$

Substituting Back into $X$

Canceling Terms

Factoring out $9$

Conclusion: $X \subseteq Y$

Final Answer: $X \cup Y$

The Sigma Insight: Types of Sets and Set Operations

The Beauty of Pattern Recognition

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If X={4n−3n−1:n∈N} and Y={9(n−1):n∈N}, where N is the set of natural numbers, then X∪Y is equal to:

Select Answer:

Visualized Solution

The Problem Statement

Analyzing Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

Elements of Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

Analyzing Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Elements of Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Hypothesis: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X⊆Y

Proving with Binomial Theorem

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+3)n

Substituting Back into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X