Question

Let $A_n={{}^{n}C}_0{{}^{n}C}_1+{{}^{n}C}_1{{}^{n}C}_2+....+{{}^{n}C}_{n-1}{{}^{n}C}_n$ and
$\frac{A_{n+1}}{A_n}=\frac{15}{4}$, then $n$ equals

• A. $8,4$
• B. $4,6$
• C. $2,4$
• D. $8,6$

Answer 1

Answer: (C)

$2,4$

$A_n={{}^{n}C}_0{{}^{n}C}_{n-1}+{{}^{n}C}_1{{}^{n}C}_{n-2}+....+{{}^{n}C}_0{{}^{n}C}_{n-1}={{}^{2n}C}_{n-1}$
Now
$\frac{A_{n+1}}{A_n}=\frac{(2n+2)!}{n!(n+2)!}\frac{(n-1)!(n+1)!}{\left(2n\right)!}=\frac{(2n+2)(2n+1)}{n(n+2)}=\frac{15}{4}$
$\Rightarrow n=2,4$