Question

If $A$ and $B$ are coefficients of $x^n$ in the expansion of${(1+x)}^{2n}$ and${(1+x)}^{2n-1}$ respectively, then$\frac{A}{B}$ is equal

• A. $4$
• B. $2$
• C. $9$
• D. $6$

Answer 1

The correct option is B

$2$

Explanation for the correct option:

Find the coefficient of${\mathit{x}}^{\mathbf{n}}\mathbf{}\mathbf{:}$

We have given,

$A$ and$B$ are coefficients of$x^n$ in the expansion of${(1+x)}^{2n}$ and${(1+x)}^{2n-1}$.

So,

$\begin{array}{rcl}A& =& {}^{2n}C_n,\\ B& =& {}^{2n-1}C_n\\ \therefore \frac{A}{B}& =& \frac{{}^{2n}C_n}{{}^{2n-1}C_n}\\ & =& \frac{{\displaystyle \raisebox{1ex}{$2n!$}\!\left/ \!\raisebox{-1ex}{$(2n-n)!n!$}\right.}}{({\displaystyle \raisebox{1ex}{$2n-1)!$}\!\left/ \!\raisebox{-1ex}{$(2n-1-n)!n!$}\right.}}\\ & =& \frac{\raisebox{1ex}{$2n!$}\!\left/ \!\raisebox{-1ex}{$\left(n\right)!n!$}\right.}{(\raisebox{1ex}{$2n-1)!$}\!\left/ \!\raisebox{-1ex}{$(n-1)!n!$}\right.}\\ & =& \frac{\raisebox{1ex}{$2n\times (2n-1)!$}\!\left/ \!\raisebox{-1ex}{$\left(n\right)(n-1)!!n!$}\right.}{(\raisebox{1ex}{$2n-1)!$}\!\left/ \!\raisebox{-1ex}{$(n-1)!n!$}\right.}\\ & =& \frac{2n}{n}\\ \frac{A}{B}& =& 2\end{array}$

Hence, option$\left(B\right)$ is correct.