Question

The $(m+1{)}^{th}$ term of ${(\frac{x}{y}+\frac{y}{x})}^{2m+1}$ is :

• A. independent of $x$
• B. a constant
• C. depends on the ration $\frac{x}{y}$ and $m$
• D. none of these

Answer 1

The correct option is D none of these

$Tr={}^n{C}_r{a}^{n-r}{b}^r$

Here $r=m+1$

$a=\frac{x}{y}$

$b=\frac{y}{x}$

$n=2m+1$

$Tr={}^{2m+1}{C}_{m+1}{\left(\frac{x}{y}\right)}^{2m+1-m-1}{\left(\frac{y}{x}\right)}^{m+1}$

$=\frac{(2m+1)!}{(m+1)!(2m+1-m-1)!}{\left(\frac{x}{y}\right)}^m{\left(\frac{y}{x}\right)}^{m+1}$

$=\frac{(2m+1)!}{(m+1)!\left(m\right)!}\left(\frac{x^{m-m-1}}{y^{m-m-1}}\right)$

$=\frac{(2m+1)!}{(m+1)!m!}\frac{y}{x}$

$\therefore \left(D\right)$.