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

The correct option is C $2,4$

${(1+x)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+...{+}^n{C}_n{x}^n$

${(1+\frac{1}{x})}^n{=}^n{C}_0{+}^n{C}_1\left(\frac{1}{x}\right){+}^n{C}_2{\left(\frac{1}{x}\right)}^2+...{+}^n{C}_n{\left(\frac{1}{x}\right)}^n$

$A_n=coefficientofxin{(1+x)}^n{(1+\frac{1}{x})}^n$

$=coefficientofxin\frac{{(1+x)}^{2n}}{x^n}$

$=coefficientof{x}^{n+1}in{(1+x)}^{2n}$

${=}^{2n}{C}_{n+1}$

$A_{n+1}{=}^{2(n+1)}{C}_{n+1+1}{=}^{2n+2}{C}_{n+2}$

$\frac{A_{n+1}}{A_n}=\frac{15}{4}$(given)

$\frac{{}^{2n+2}{C}_{n+2}}{{}^{2n}{C}_{n+1}}=\frac{15}{4}$

$\Rightarrow \frac{(2n+2)!}{(n+2)!n!}\times \frac{(n+1)!(n-1)!}{\left(2n\right)!}=\frac{15}{4}$

$\Rightarrow \frac{(2n+2)(2n+1)}{n(n+2)}=\frac{15}{4}$

$\Rightarrow 16n^2+24n+8=15n^2+30n$

$\Rightarrow n^2-6n+8=0$

$\Rightarrow n^2-4n-2n+8=0$

$\Rightarrow n(n-4)-2(n-4)=0$

$\Rightarrow (n-4)(n-2)=0$

$\therefore n=2,4$