Question

Let $A_k=\frac{{{}^{n}C}_k}{{{}^{n}C}_k+{{}^{n}C}_{k+1}}$. If $\sqrt[3]{3{\sum }_{k=0}^n{A}_k}=4$ then $n$ equals

• A. $122$
• B. $124$
• C. $126$
• D. $128$

Answer 1

The correct option is C $126$

$\frac{{}^n{C}_k}{{}^n{C}_k+{}^n{C}_{k+1}}$
$=\frac{{}^n{C}_k}{{}^{n+1}{C}_{k+1}}$
$=\frac{n!(k+1)!.(n-k)!}{(n+1)!.k!(n-k)!}$
$=\frac{k+1}{n+1}$
Form $k=0$ to $k=n$ will be
$=\frac{n+1}{n+1}+\frac{n(n+1}{2(n+1)}$
$=\frac{(n+1)(n+2)}{2(n+1)}$
$=\frac{n+2}{2}$
$=64$
Hence
$n+2=128$
$n=126$