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We already know that we can represent this binomial as the following:

$$(a+b)^K=\sum _{n=0}^K \binom{K}{n} b^n a^{K-n};$$

where $\binom{K}{n} = \frac{K!}{n! (K-n)!}$

I want to know if this representation is correct for the reciprocal binomial expression:

$$\frac{1}{(a+b)^K}=\sum _{n=0}^K \frac{1}{\binom{K}{n} b^n a^{K-n}};$$

where $K$ is a positive integer larger than zero.

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2 Answers

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Absolutely not. What you wrote is equivalent to claiming that $$\frac{1}{a+b} = \frac{1}{a} + \frac{1}{b}.$$

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If $|a|>|b|$ then

$$ \begin{align} & \frac{1}{(a+b)^K} \\[8pt] = {} & (a+b)^{-K} = a^{-K} + (-K)a^{-K-1}b + \frac{-K(-K-1)}{2} a^{-K-1}b^2 \\[8pt] & {} + \frac{-K(-K-1)(-K-2)}{6} a^{-K-3} b^3 + \cdots \\[8pt] = {} & \sum_{n=0}^\infty \binom{-K}{n} a^{-K-n} b^n. \end{align} $$

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