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If $|\mathrm{x}| < \frac{1}{2}$, what is the value of
$1+n\left[\frac{x}{1-x}\right]+\left[\frac{n(n+1)}{2 !}\right]\left[\frac{x}{1-x}\right]^{2}+\ldots \ldots \ldots \ldots \ldots \infty ?$
Options:
$1+n\left[\frac{x}{1-x}\right]+\left[\frac{n(n+1)}{2 !}\right]\left[\frac{x}{1-x}\right]^{2}+\ldots \ldots \ldots \ldots \ldots \infty ?$
Solution:
1826 Upvotes
Verified Answer
The correct answer is:
$\left[\frac{1-x}{1-2 x}\right]^{n}$
Given that $1+\mathrm{n}\left[\frac{\mathrm{x}}{1-\mathrm{x}}\right]+\frac{\mathrm{n}(\mathrm{n}+1)}{2 !}\left[\frac{\mathrm{x}}{1-\mathrm{x}}\right]^{2}+\ldots \infty$ is
expansion of $\left[1-\frac{x}{1-x}\right]^{-n}$.
So, it is $=\left[1-\frac{x}{1-x}\right]^{-n}=\left[\frac{1-x-x}{1-x}\right]^{-n}=\left[\frac{1-x}{1-2 x}\right]^{n}$
expansion of $\left[1-\frac{x}{1-x}\right]^{-n}$.
So, it is $=\left[1-\frac{x}{1-x}\right]^{-n}=\left[\frac{1-x-x}{1-x}\right]^{-n}=\left[\frac{1-x}{1-2 x}\right]^{n}$
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