C r n - 1 = k 2 - 8 C r + 1 n if and only if :

Given: ${}^{n-1}C_r=\left(k^2-8\right){}^{n}C_{r+1} ...\left(i\right)$

$\Rightarrow \underset{r\ge 0}{\underbrace{r+1\ge 0, r\ge 0}}$

$\Rightarrow \frac{{}^{n-1}C_r}{{}^{n}C_{r+1}}=k^2-8$

$\Rightarrow \frac{{\displaystyle \frac{\left(n-1\right)!}{r!\left(n-r-1\right)!}}}{{\displaystyle \frac{n!}{\left(r+1\right)!\left(n-r-1\right)!}}}=k^2-8$

$\Rightarrow \frac{r+1}{n}=k^2-8$

$\Rightarrow k^2-8>0$

$\Rightarrow (k-2\sqrt{2})(k+2\sqrt{2})>0$

$\Rightarrow k\in (-\infty ,-2\sqrt{2})\cup (2\sqrt{2},\infty ) ...\left(ii\right)$

Using equation $\left(i\right)$,

$\Rightarrow n\ge r+1$

$\Rightarrow \frac{r+1}{n}\le 1$

$\Rightarrow k^2-8\le 1$

$\Rightarrow k^2-9\le 0$

$\Rightarrow -3\le \mathrm{k}\le 3 ...\left(ii\right)$

From equation $\left(ii\right) \text{and} \left(iii\right)$ we get,

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