Question

If ${21}^{st}$ and ${22}^{nd}$ terms in the expansion of $(1+x{)}^{44}$ are equal, then $x$ is equal to

Answer 1

To find the value of $\mathit{x}$:

The $r^{th}$ term of the expansion $(1+x{)}^{44}$ is $\left(\begin{array}{c}n\\ r-1\end{array}\right)a^{n-(r-1)}{b}^{r-1}$.

Let $T_r$ denote the $r^{th}$term. Here, $n=44,a=1,b=x$

$T_{21}=\left(\begin{array}{c}44\\ 20\end{array}\right)1^{44-20}{x}^{20}$

$T_{22}=\left(\begin{array}{c}44\\ 21\end{array}\right)1^{44-21}{x}^{21}$

It is given that $T_{21}=T_{22}$.

$\begin{array}{rcl}\frac{T_{22}}{T_{21}}& =& \frac{\left(\begin{array}{c}44\\ 21\end{array}\right)1^{44-21}{x}^{21}}{\left(\begin{array}{c}44\\ 20\end{array}\right)1^{44-20}{x}^{20}}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{}\left(\begin{array}{c}n\\ r\end{array}\right)\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{}\mathbf{r}\mathbf{!}}\mathbf{]}\\ 1& =& \frac{24x}{21}\\ 21& =& 24x\\ x& =& \frac{21}{24}\\ \therefore x& =& \frac{7}{8}\end{array}$

Hence, value of $x=\frac{7}{8}$.