The sum of 1 + n(1 - frac{1}{x}) + frac{n(n + | vedclass

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(A) The general 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 wi

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$x^n$

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(A) The general 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 $1 + n(1 - \frac{1}{x}) + \frac{n(n + 1)}{2!}(1 - \frac{1}{x})^2 + \dots \infty$,we identify $y = (1 - \frac{1}{x})$. Substituting this into the formula,we get the sum as $(1 - (1 - \frac{1}{x}))^{-n}$. Simplifying the expression inside the bracket: $1 - 1 + \frac{1}{x} = \frac{1}{x}$. Thus,the sum is $(\frac{1}{x})^{-n} = (x^{-1})^{-n} = x^n$.

The sum of $1 + n(1 - \frac{1}{x}) + \frac{n(n + 1)}{2!}(1 - \frac{1}{x})^2 + \dots \infty$ is:

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