^n-1C_r=(k^2-8)^nC_r+1 if and only if :

Question

${}^{n-1}{C}_{r}=({k}^{2}-8){}^{n}{C}_{r+1}$ if and only if :

Accepted Answer

$${ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}}=(\mathrm{k}^2-8){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$$

$$\underbrace{\mathrm{r}+1 \geq 0, \quad \mathrm{r} \geq 0}_{\mathrm{r} \geq 0}$$

$$\begin{aligned}
& \frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8 \
& \frac{r+1}{n}=k^2-8 \
& \Rightarrow k^2-8>0 \
& (k-2 \sqrt{2})(k+2 \sqrt{2})>0
\end{aligned}$$

$$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)\quad \text{.... (I)}$$

$$\begin{aligned}
\therefore & \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1 \
\Rightarrow & \mathrm{k}^2-8 \leq 1 \
& \mathrm{k}^2-9 \leq 0 \
& -3 \leq \mathrm{k} \leq 3 \quad \text{.... (II)}
\end{aligned}$$

From equation (I) and (II) we get

$$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$$

Solutions · 1