^{n - 1}C_r = (k^2 - 3) cdot ^nC_{r + 1} if k | vedclass

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(D) Given the equation: $^{n - 1}C_r = (k^2 - 3) \cdot ^nC_{r + 1}$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we expand both sides:$\frac{(n - 1

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$(\sqrt{3}, 2)$

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(D) Given the equation: $^{n - 1}C_r = (k^2 - 3) \cdot ^nC_{r + 1}$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we expand both sides:$\frac{(n - 1)!}{r!(n - r - 1)!} = (k^2 - 3) \cdot \frac{n!}{(r + 1)!(n - r - 1)!}$.Canceling common terms $(n - r - 1)!$ and $(n - 1)!$ from both sides:$1 = (k^2 - 3) \cdot \frac{n}{r + 1}$.Rearranging for $k^2$:$k^2 - 3 = \frac{r + 1}{n} \Rightarrow k^2 = \frac{r + 1}{n} + 3$.Since $0 \le r \le n - 1$,we have $1 \le r + 1 \le n$,which implies $\frac{1}{n} \le \frac{r + 1}{n} \le 1$.Thus,$k^2 \in [\frac{1}{n} + 3, 4]$.For $n \ge 2$,the range of $k^2$ is $(3, 4]$.Taking the square root,$k \in [-2, -\sqrt{3}) \cup (\sqrt{3}, 2]$.Comparing with the given options,the interval $(\sqrt{3}, 2)$ is the correct subset.

$^{n - 1}C_r = (k^2 - 3) \cdot ^nC_{r + 1}$ if $k \in$

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