Question

If ${}^{n-1}{C}_r=(k^2-3{)}^n{C}_{r+1}$, then $k\in$

• A. $(\infty ,-2)$
• B. $(2,\infty )$
• C. $[-\sqrt{3},\sqrt{3}]$
• D. $(\sqrt{3},2]$

Answer 1

The correct option is D $(\sqrt{3},2]$

$k^2-3=\frac{{}^{n-1}{C}_r}{{}^n{C}_{r+1}}$
$=\frac{(n-1)!}{r!.(n-r-1)!}\times \frac{(r+1)!(n-r-1)!}{n!}$
$=\frac{r+1}{n}$
$\therefore k^2=\frac{r+1}{n}+3$
Also $n-1\ge r\Rightarrow n\ge r+1(r\ge 0)$
$k^2\in (3,4]\Rightarrow k\in (\sqrt{3},2]$