First slide

Binomial theorem for negative integral and rational Index

Question

If $|\frac{2 x}{1 + x}| < 1$ , then $1 + n (\frac{2 x}{1 + x}) + \frac{n (n + 1)}{2!} {(\frac{2 x}{1 + x})}^{2} + \dots$ is equal to

Easy

A

${(\frac{2 x}{1 + x})}^{n}$
B

${(\frac{1 + x}{2 x})}^{n}$
C

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

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

Solution

Required value is
${(1 - \frac{2 x}{1 + x})}^{- n} = {(\frac{1 + x - 2 x}{1 + x})}^{- n} = {(\frac{1 - x}{1 + x})}^{- n} = {(\frac{1 + x}{1 - x})}^{n}$

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