Question

The term independent of $x$ in the expansion of ${\left(\left[\left(t^{-1}-1\right)x+{\left(t^{-1}+1\right)}^{-1}{x}^{-1}\right]\right)}^8$ is :

• A. $56{\left(\frac{1-t}{1+t}\right)}^3$
• B. $56{\left(\frac{1+t}{1-t}\right)}^3$
• C. $70{\left(\frac{1-t}{1+t}\right)}^4$
• D. $70{\left(\frac{1+t}{1-t}\right)}^4$

Answer 1

Answer: (C)

$\begin{array}{l}{\left[\left(t^{-1}-1\right)x+{\left(t^{-1}+1\right)}^{-1}{x}^{-1}\right]}^8\\ ={\left[\left(\frac{1}{t}-1\right)x+{\left(\frac{1}{t}+1\right)}^{-1}\frac{1}{x}\right]}^8\end{array}$
Let $T_{r+1}$ be the term independent of $x$, then
$\begin{array}{l}{T}_{r+1}={}^8{C}_r{\left(\frac{1}{t}-1\right)}^{8-r}{x}^{8-r}\cdot {\left(\frac{1}{t}+1\right)}^{-r}{\left(\frac{1}{x}\right)}^r\\ ={}^8{C}_r{\left(\frac{1}{t}-1\right)}^{8-r}\cdot {\left(\frac{1}{t}+1\right)}^{-r}\cdot x^{8-2r}\\ \therefore 8-2r=0\Rightarrow r=4\end{array}$
$\therefore T_5$ is the term independent of $x$ and
$\begin{array}{l}{T}_5={}^8{C}_4{\left(\frac{1}{t}-1\right)}^4\cdot {\left(\frac{1}{t}+1\right)}^{-4}\\ ={}^8{C}_4{\left(\frac{1-t}{t}\right)}^4\cdot {\left(\frac{1+t}{t}\right)}^{-4}\\ ={}^8{C}_4{\left(\frac{1-t}{1+t}\right)}^4=\frac{8.7\cdot 6.5}{4.3\cdot 2.1}{\left(\frac{1-t}{1+t}\right)}^4\\ =70\cdot {\left(\frac{1-t}{1+t}\right)}^4\end{array}$