Question

If $T_r$ denotes the $r^{th}$ term in the expansion of ${(x+\frac{1}{x})}^{23}$, then

• A. $T_{12}=T_{13}$
• B. $x^2.T_{13}=T_{12}$
• C. $x^2.T_{12}=T_{13}$
• D. $T_{12}+T_{13}=25$

Answer 1

The correct option is B $x^2.T_{13}=T_{12}$

We know that, general term expression for $(a+b{)}^n$
$T_{r+1}{=}^n{C}_r\times a^n\times {\left(b\right)}^{n-r}$
$T_{12}{=}^{23}{C}_{11}\times x^{12}\times {\left(\frac{1}{x}\right)}^{11}$
$T_{12}{=}^{23}{C}_{11}x$
$T_{13}{=}^{23}{C}_{12}\times x^{11}\times {\left(\frac{1}{x}\right)}^{12}$
$T_{13}{=}^{23}{C}_{12}\left(\frac{1}{x}\right)$
${}^{23}{C}_{12}{=}^{23}{C}_{11}$ ....as $(11+12=23)$
$T_{12}{=}^{23}{C}_{11}{x}^2\times \frac{1}{x}$
$T_{12}=x^2{\times }^{23}{C}_{12}\times \frac{1}{x}$

$T_{12}=x^2\times T_{13}$

Hence, option B is the correct answer.