First slide

Binomial theorem for positive integral Index

Question

If $^{n - 1} C_{r} = {(k^{2} - 3)}^{n} C_{r + 1},$ then k is belongs to

Moderate

A

$(- \infty, - 2]$
B

$[2, \infty)$
C

$[- \sqrt{3}, \sqrt{3}]$
D

$(\sqrt{3}, 2]$

Solution

$∵ (k^{2} - 3) = \frac{^{n - 1} C_{r}}{^{n} C_{r + 1}} = \frac{^{n - 1} C_{r}}{{(\frac{n}{r + 1})}^{n - 1} C_{r}} = (\frac{r + 1}{n}) - - - - - (i)$
$∵ 0 \leq r \leq n - 1$
$\Rightarrow 1 \leq r + 1 \leq n \Rightarrow \frac{1}{n} \leq \frac{r + 1}{n} \leq 1$
$\Rightarrow \frac{1}{n} \leq k^{2} - 3 \leq 1$
$\Rightarrow 3 + \frac{1}{n} \leq k^{2} \leq 4$
When $n \to \infty, 3 < k^{2} < 4 or k \in [- 2, - \sqrt{3}] \cup (\sqrt{3}, 2]$

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