Question

If $-\frac{1}{3}<x<1$ then $1+n\left(\frac{2x}{1+x}\right)+\frac{n(n+1)}{2!}{\left(\frac{2x}{1+x}\right)}^2+.......$ is equal to

• A. ${\left(\frac{2x}{1+x}\right)}^n$
• B. ${\left(\frac{1+x}{2x}\right)}^n$
• C. ${\left(\frac{1-x}{1+x}\right)}^n$
• D. ${\left(\frac{1+x}{1-x}\right)}^n$

Answer 1

The correct option is D ${\left(\frac{1+x}{1-x}\right)}^n$

$\text{Using, expansion of}(1-x{)}^{-n}$

$(1-x{)}^{-n}=1+nx+\frac{n(n+1)}{2}{x}^2+\frac{n(n+1)(n+2)x^3+\dots .}{3}$

$\text{(To be Remembered)}$

$\text{on comparing with given equation}$

$x=\frac{2x}{1+x}$

$\text{so, it becomes}$

$\text{Let}S=1+n\left(\frac{2x}{1+x}\right)+\frac{n(n+1)}{2}{\left(\frac{2x}{1+x}\right)}^2+\dots \dots$

$S={(1-\frac{2x}{1+x})}^{-n}$

$={\left(\frac{1-x}{1+x}\right)}^{-n}$

$={\left(\frac{1+x}{1-x}\right)}^n$

$\text{So, the correct answer is (D)}$