If ^{n-1}C_r = (k^2 - 3) cdot ^nC_{r+1},then | 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 have: $\frac{(n-1)!}{r!(n-r-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 have: $\frac{(n-1)!}{r!(n-r-1)!} = (k^2 - 3) \cdot \frac{n!}{(r+1)!(n-r-1)!}$ Simplifying the factorials: $\frac{1}{r!} = (k^2 - 3) \cdot \frac{n}{(r+1)r!}$ $1 = (k^2 - 3) \cdot \frac{n}{r+1}$ $k^2 - 3 = \frac{r+1}{n}$ $k^2 = \frac{r+1}{n} + 3$ 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 value of $\frac{1}{n} + 3$ lies in the interval $(3, 3.5]$. Therefore,$k \in [-2, -\sqrt{\frac{1}{n} + 3}] \cup [\sqrt{\frac{1}{n} + 3}, 2]$. Comparing with the given options,the interval $(\sqrt{3}, 2)$ is the most appropriate subset.

If $^{n-1}C_r = (k^2 - 3) \cdot ^nC_{r+1}$,then the range of $k$ is:

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