If n-1Cr = (k² - 3) ⁿ Cr+1, then k ∈ (a) (-∞, -2] ∪ [2, ∞) (b) [2, ∞) (c) [-√3, √3] (d) (√3, 2]

Q: Wait, why do we reject the negative root here? Or do we?

That is a fantastic question! We actually do not reject the negative root. When we solve $3 < k^2 \le 4$, we are looking for all values of $k$ whose square falls in that range. Since $(-2)^2 = 4$ and $(-\sqrt{3})^2 = 3$, any value between $-2$ and $-\sqrt{3}$ is a valid solution. Always remember that squaring a negative number yields a positive result, so the negative interval is just as valid as the positive one!

Q: Why is the range of $r+1$ between $1$ and $n$?

Great catch. For $^nC_{r+1}$ to be defined, the lower index must be at least $0$, so $r+1 \ge 0$. However, since $r$ is a non-negative integer (as it is an index in a combination), $r \ge 0$, which implies $r+1 \ge 1$. Combined with the upper bound $r+1 \le n$, we get $1 \le r+1 \le n$. This is the crucial step that bridges the gap between combinatorics and inequalities.

Q: Why can't we just solve for $k$ directly? Why do we need to use inequalities?

Because $r$ and $n$ are variables, not fixed numbers. The ratio $\frac{r+1}{n}$ is not a single value; it is a range of possible values depending on the choice of $r$ and $n$. We are not solving for a specific $k$ that works for one $n$; we are finding the range of $k$ that allows the equation to hold true for *some* valid $n$ and $r$. The inequality approach is the only way to capture all those possibilities.

Question

If $^{n-1}C_r = (k^2 - 3) ^nC_{r+1}$, then $k \in$

Options:

(A) $(-\infty, -2] \cup [2, \infty)$

(B) $[2, \infty)$

(C) $[-\sqrt{3}, \sqrt{3}]$

(D) $(\sqrt{3}, 2]$

Sigma Percentile · Accepted Answer

### The Beauty of Algebraic Symmetry: A Journey into Combinatorics

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of factorials and constants.

We are given the equation $^{n-1}C_r = (k^2 - 3) ^nC_{r+1}$ and asked to find the range of $k$. It looks intimidating, but I promise you, there is a hidden elegance here waiting to be uncovered. Let us take a breath and dive in.

### Phase 1: The Algebraic Setup

Our first instinct should always be to simplify. We have a constant $k$ trapped inside a term $(k^2 - 3)$. Let us isolate it by rearranging the equation:

$$k^2 - 3 = \frac{^{n-1}C_r}{^nC_{r+1}}$$

Now, we must confront the binomial coefficients. Recall the definition: $^nC_r = \frac{n!}{r!(n-r)!}$. This is the heartbeat of combinatorics.

Let us expand both the numerator and the denominator with this definition:

$$^{n-1}C_r = \frac{(n-1)!}{r!(n-1-r)!}$$

$$^nC_{r+1} = \frac{n!}{(r+1)!(n-r-1)!}$$

When we substitute these into our ratio, we get a complex-looking fraction. But do not panic! Dividing by a fraction is just multiplying by its reciprocal. Let us flip that denominator and multiply:

$$k^2 - 3 = \frac{(n-1)!}{r!(n-1-r)!} \cdot \frac{(r+1)!(n-r-1)!}{n!}$$

### Phase 2: The Elegance of Cancellation

This is the moment of truth. Look at the terms. We have $(n-1)!$ in the numerator and $n!$ in the denominator. Since $n! = n \cdot (n-1)!$, they will cancel beautifully.

Similarly, $(r+1)! = (r+1) \cdot r!$, so the $r!$ terms will vanish. And notice $(n-r-1)!$ appears in both the numerator and denominator? They cancel out entirely!

After the dust settles, we are left with something remarkably simple:

$$k^2 - 3 = \frac{r+1}{n}$$

Isn't that satisfying? All that complexity collapsed into a simple ratio. But we are not done yet; we need to find the range of $k$ by understanding the constraints on $r$ and $n$.

### Phase 3: The Constraint Trap

In the world of combinations, $r$ and $n$ are not just any numbers. For $^{n-1}C_r$ to exist, we must have $0 \le r \le n-1$. For $^nC_{r+1}$ to exist, we must have $0 \le r+1 \le n$.

Combining these, we find that $1 \le r+1 \le n$. This is the bridge we need. If we divide this inequality by $n$ (which is a positive integer), we get:

$$\frac{1}{n} \le \frac{r+1}{n} \le 1$$

Since $n \ge 1$, the term $\frac{1}{n}$ is strictly greater than $0$. Thus, our ratio $\frac{r+1}{n}$ is trapped between $0$ and $1$. Specifically, $0 < \frac{r+1}{n} \le 1$.

Since this ratio is equal to $k^2 - 3$, we have our inequality:

$$0 < k^2 - 3 \le 1$$

### Phase 4: The Final Reveal

Now, we are in the home stretch. To isolate $k^2$, we add $3$ to all parts of the inequality:

$$3 < k^2 \le 4$$

This is the critical moment where many students stumble. We are solving for $k$, not $k^2$. When we take the square root, we must remember that $\sqrt{k^2} = |k|$.

So, we have:

$$\sqrt{3} < |k| \le 2$$

This absolute value inequality splits into two distinct regions on the number line. For the positive side, we have $\sqrt{3} < k \le 2$. For the negative side, we have $-2 \le k < -\sqrt{3}$.

Combining these, our solution is **$k \in [-2, -\sqrt{3}) \cup (\sqrt{3}, 2]$**. You have navigated the algebra, respected the constraints, and arrived at the truth. Well done!

Animated Solution for Mathematics - Permutations and Combinations: If $^{n - 1} C_{r} = (k^{2} - 3)^{n} C_{r + 1}$ , then $k \in$

Select Answer:

Visualized Solution

The Equation and Goal

Expanding the Combinations

Setting up the Ratio

Simplifying the Factorials

Constraints on $r$ and $n$

Bounding the Ratio

Solving for $k^{2}$

Solving for $k$ (Positive and Negative Roots)

Final Intervals for $k$

Matching with Options

The Sigma Insight: Combinations and Selection

The Beauty of Algebraic Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Permutations and Combinations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If n−1Cr​=(k2−3)nCr+1​, then k∈