જો { }^{n-1} C_r=left(k^2-8right){ }^n C_{r+1 | vedclass

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Question

$ { }^{n-1} C_r=(k^2-8){ }^n C_{T+1} $

$\underbrace{r+1 \geq 0, \quad r \geq 0}_{r \geq 0}$

$\frac{{ }^{n-1} C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}

Vedclass · Accepted Answer

$2 \sqrt{2}<\mathrm{k} \leq 3$

Vedclass · Answer

$ { }^{n-1} C_r=(k^2-8){ }^n C_{T+1} $

$\underbrace{r+1 \geq 0, \quad r \geq 0}_{r \geq 0}$

$\frac{{ }^{n-1} C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}+1}}=\mathrm{k}^2-8$

$\frac{\mathrm{r}+1}{\mathrm{n}}=\mathrm{k}^2-8$

$\Rightarrow \mathrm{k}^2-8>0$

$(\mathrm{k}-2 \sqrt{2})(\mathrm{k}+2 \sqrt{2})>0$

$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)$   $......(I)$

$	herefore \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 $                              $......(II)$

From equation $(I)$ and $(II)$ we get

$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$

જો ${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$ તો અને તો જ

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