Question

Let M(n) be the largest integer in m such that ${}^m{C}_{n-1}{>}^{m-1}{C}_n,$ then the value of $\underset{n\to \infty }{lim}$ $\frac{M\left(n\right)}{n}$ is

• A. $\frac{3-\sqrt{5}}{2}$
• B. $\frac{3+\sqrt{5}}{2}$
• C. $\frac{\sqrt{5}-1}{2}$
• D. $\frac{\sqrt{5}+1}{2}$

Answer 1

The correct option is B

$\frac{3+\sqrt{5}}{2}$

$\frac{m!}{(n-1)!(m-n+1)!}>\frac{(m-1)!}{n!(m-1-n)!}\Rightarrow \frac{m}{(m-n+1)(m-n)}>\frac{1}{n}\Rightarrow mm>(m-n+1)(m-n)\Rightarrow m^2-3mn+m+n^2-n<0\Rightarrow m^2-(3n-1)m+(n^2-n)<0\therefore mϵ(\frac{3n-1-\sqrt{5n^2-2n+1}}{2},\frac{3n-1+\sqrt{5n^2-2n+1}}{2})$
Clearly some integer must be lying in this internal, let it be M(n)
$\therefore \frac{3n-1-\sqrt{5n^2-2n+1}}{2}-1M\left(n\right)<\frac{3n-1+\sqrt{5n^2-2n+1}}{2}$
$\therefore$ According to sandwich theorem
$\underset{n\to \infty }{lim}\frac{M\left(n\right)}{n}=\frac{3+\sqrt{5}}{2}$