If |x|

Question

(A) The binomial expansion for a negative index is given by $(1 - y)^{-n} = 1 + ny + \frac{n(n + 1)}{2!}y^2 + \dots \infty$.Comparing this with the gi

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

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(A) The binomial expansion for a negative index is given by $(1 - y)^{-n} = 1 + ny + \frac{n(n + 1)}{2!}y^2 + \dots \infty$.Comparing this with the given series $1 + n\left( \frac{2x}{1 + x} \right) + \frac{n(n + 1)}{2!}\left( \frac{2x}{1 + x} \right)^2 + \dots \infty$,we identify $y = \frac{2x}{1 + x}$.Thus,the sum is $\left( 1 - \frac{2x}{1 + x} \right)^{-n}$.Simplifying the expression inside the bracket: $1 - \frac{2x}{1 + x} = \frac{1 + x - 2x}{1 + x} = \frac{1 - x}{1 + x}$.Therefore,the sum is $\left( \frac{1 - x}{1 + x} \right)^{-n} = \left( \frac{1 + x}{1 - x} \right)^n$.

If $|x| < 1$,then the value of $1 + n\left( \frac{2x}{1 + x} \right) + \frac{n(n + 1)}{2!}\left( \frac{2x}{1 + x} \right)^2 + \dots \infty$ will be

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