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

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

$⇒ \mathop{\mathop{r + 1 \ge 0, r \ge 0}_{⏟}}_{r \ge 0}$

$⇒ \dfrac{Cr n - 1}{Cr + 1 n} = {k}^{2} - 8$

$⇒ \dfrac{\dfrac{\left(n - 1\right) !}{r ! \left(n - r - 1\right) !}}{\dfrac{n !}{\left(r + 1\right) ! \left(n - r - 1\right) !}} = {k}^{2} - 8$

$⇒ \dfrac{r + 1}{n} = {k}^{2} - 8$

$⇒ {k}^{2} - 8 > 0$

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

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

Using equation \left(i\right)$,</p><p>$⇒ n \ge r + 1$</p><p>$⇒ \dfrac{r + 1}{n} \le 1$</p><p>$⇒ {k}^{2} - 8 \le 1$</p><p>$⇒ {k}^{2} - 9 \le 0$</p><p>$⇒ - 3 \le k \le 3 . . . ( ii )$</p><p>From equation \left(ii\right)\text{and}\left(iii\right)$ we get,

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