Question

If $T_r$ denotes the $r^{th}$ term in the expansion of ${(x+\frac{1}{x})}^{25}$ , then

• A. $T_{12}=T_{13}$
• B. $\frac{15}{11x^2}\cdot T_{11}=T_{12}$
• C. $\frac{15}{23x^2}\cdot T_{12}=T_{13}$
• D. $T_{12}+T_{13}=25$

Answer 1

The correct option is B $\frac{15}{11x^2}\cdot T_{11}=T_{12}$

Given, ${(x+\frac{1}{x})}^{25}$

Now, $T_{r+1}={}^{25}{C}_r{x}^{25-r}\times {\left(\frac{1}{x}\right)}^r$

$={}^{25}{C}_r{x}^{25-2r}$

So, $T_{12}={}^{25}{C}_{11}\times x^{25-2\times 11}$
$={}^{25}{C}_{11}\times x^3$ ........... $\left(1\right)$

$T_{13}={}^{25}{C}_{12}\times x^{25-2\times 12}$

$={}^{25}{C}_{12}\times x$ ............ $\left(2\right)$

$T_{11}={}^{25}{C}_{10}\times x^5$ ............ $\left(3\right)$

From (1) & (2)

$\frac{T_{12}}{T_{13}}=\frac{{}^{25}{C}_{11}\times x^3}{{}^{25}{C}_{12}\times x}=\frac{\frac{25!}{11!14!}\times x^3}{\frac{25!}{12!13!}\times x}=\frac{12}{14}{x}^2$

$\Rightarrow T_{13}=\frac{7}{6x^2}{T}_{12}$

From $\left(3\right)$ & $\left(1\right)$ , we get

$\frac{T_{11}}{T_{12}}=\frac{{}^{25}{C}_{10}\times x^5}{{}^{25}{C}_{11}\times x^3}=\frac{\frac{25!}{10!15!}\times x^5}{\frac{25!}{11!14!}\times x^3}=\frac{11}{15}{x}^2$
$\Rightarrow T_{12}=\frac{15}{11x^2}{T}_{11}$
Hence, Option (B) is the correct answer.