If ^n-1C_r = (k² - 8)ⁿC_r+1 then

The correct answer is:
$$k\in [-3,-2\sqrt{2})\cup (2\sqrt{2},3]$$

Step-by-Step Derivation and Explanation:

Step 1: Understand the given relation
We are given the relation:
$${}^{n-1}{C}_r=(k^2-8){}^n{C}_{r+1}$$

Step 2: Express the combination terms in a simplified form
Using the properties of combinations, we know that:
$${}^n{C}_{r+1}=\frac{n}{r+1}\times {}^{n-1}{C}_r$$
Substituting this relation back into our original equation:
$${}^{n-1}{C}_r=(k^2-8)\times \frac{n}{r+1}\times {}^{n-1}{C}_r$$

Since ${}^{n-1}{C}_r\ne 0$, we can divide both sides by ${}^{n-1}{C}_r$:
$$1=(k^2-8)\times \frac{n}{r+1}$$
Rearranging the terms, we get:
$$\frac{1}{k^2-8}=\frac{n}{r+1}$$

Step 3: Analyze the mathematical constraints on $n$ and $r$
For the combination term ${}^n{C}_{r+1}$ to be defined, the upper index must be greater than or equal to the lower index, and both must be non-negative integers:
$$n\ge r+1$$
Since $n$ and $r+1$ are positive integers:
$$\frac{n}{r+1}\ge 1$$

Substituting this constraint back into our simplified relation, we obtain the inequality:
$$\frac{1}{k^2-8}\ge 1$$

Step 4: Solve the inequality for $k$
Subtract 1 from both sides:
$$\frac{1}{k^2-8}-1\ge 0$$
Combine into a single fraction:
$$\frac{1-(k^2-8)}{k^2-8}\ge 0$$
$$\frac{9-k^2}{k^2-8}\ge 0$$
Multiply the numerator by $-1$ and reverse the inequality sign:
$$\frac{k^2-9}{k^2-8}\le 0$$

Factor the terms in the numerator and the denominator:
$$\frac{(k-3)(k+3)}{(k-2\sqrt{2})(k+2\sqrt{2})}\le 0$$

Step 5: Apply the interval sign method (Wavy Curve method)
The critical values that make the expression equal to zero or undefined are:
• Numerator zeros: $k=-3,3$ (included in the solution because of $\le$)
• Denominator zeros: $k=-2\sqrt{2},2\sqrt{2}$ (excluded from the solution as the denominator cannot be zero)

Arranging these critical points on the real number line:
$$-3<-2\sqrt{2}<2\sqrt{2}<3$$

Testing the signs in the intervals:
• For $k^2>9$ (i.e., $k<-3$ or $k>3$), the expression is positive (+).
• For $8<k^2\le 9$ (i.e., $k\in [-3,-2\sqrt{2})\cup (2\sqrt{2},3]$), the expression is negative or zero (-).
• For $k^2<8$ (i.e., $k\in (-2\sqrt{2},2\sqrt{2})$), the expression is positive (+) because both numerator and denominator are negative.

Since we require the expression to be less than or equal to zero ($\le 0$), the solution set is:
$$k\in [-3,-2\sqrt{2})\cup (2\sqrt{2},3]$$