Binomial Theorem
Combinatorial Inequality / Pascal's Identity
nta_pyq_2024_jan
Grade 11
Question:
${}^{n-1}C_r = (k^2 - 8)\,{}^{n}C_{r+1}$ if and only if:
$2\sqrt{2} < k \le 3$
$2\sqrt{3} < k \le 3\sqrt{2}$
$2\sqrt{3} < k < 3\sqrt{3}$
$2\sqrt{2} < k < 2\sqrt{3}$
Step-by-Step Solution
Key Concept: Write $\frac{{}^nC_r}{{}^nC_{r+1}} = k^2-8$, which gives $\frac{r+1}{n} = k^2-8$. Apply two constraints: (i) $k^2-8>0$ (ratio must be positive) and (ii) $\frac{r+1}{n}\le 1$ i.e. $k^2-8\le 1$.
${}^{n-1}C_r = (k^2-8){}^nC_{r+1}$ ⟹ $\frac{{}^nC_r}{{}^nC_{r+1}} = k^2-8$ ⟹ $\frac{r+1}{n}=k^2-8$.
Condition (I): $k^2-8>0$ ⟹ $k\in(-\infty,-2\sqrt{2})\cup(2\sqrt{2},\infty)$.
Condition (II): $n\ge r+1$ ⟹ $k^2-8\le1$ ⟹ $k^2\le9$ ⟹ $k\in[-3,3]$.
Intersection: $k\in[-3,-2\sqrt{2})\cup(2\sqrt{2},3]$. For positive $k$: $2\sqrt{2}<k\le3$.
Correct Answer: 1