Question

If $\left|x\right|<1$, 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

Answer: (D)

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

$1+n\left(\frac{2x}{1+x}\right)+\frac{n(n+1)}{2!}(\frac{2x}{1+x}{)}^2...\infty$
Now this is infinite binomial expansion and can be written as
$(\frac{1}{1-y}{)}^n$
$=(\frac{1}{1-\frac{2x}{1+x}}{)}^n$
$=(\frac{1+x}{1+x-2x}{)}^n$
$=(\frac{1+x}{1-x}{)}^n$