${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$ if and only if :

<p>$${ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}}=(\mathrm{k}^2-8){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$$</p> <p>$\underbrace{\mathrm{r}+1 \geq 0, \quad \mathrm{r} \geq 0}_{\mathrm{r} \geq 0}$</p> <p>$$\begin{aligned} & \frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8 \\ & \frac{r+1}{n}=k^2-8 \\ & \Rightarrow k^2-8>0 \\ & (k-2 \sqrt{2})(k+2 \sqrt{2})>0 \end{aligned}$$</p> <p>$$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)\quad \text{.... (I)}$$</p> <p>$$\begin{aligned} \therefore & \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1 \\ \Rightarrow & \mathrm{k}^2-8 \leq 1 \\ & \mathrm{k}^2-9 \leq 0 \\ & -3 \leq \mathrm{k} \leq 3 \quad \text{.... (II)} \end{aligned}$$</p> <p>From equation (I) and (II) we get</p> <p>$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$</p>