Why Does This Summation Simplify to a Power of p?

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SUMMARY

The summation $\sum_{x=0}^{\infty} \binom{x+r-2}{r-2}(1-p)^{x}$ simplifies to $p^{1-r}$ through the application of the binomial series. By substituting $x = -(1-p)$ and $\alpha = 1-r$ into the binomial series formula, the relationship is established. This simplification is crucial for understanding the convergence properties of the series in relation to the parameter $p$.

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  • Understanding of binomial coefficients, specifically $\binom{x+r-2}{r-2}$.
  • Familiarity with the binomial series expansion, $(1+x)^\alpha = \sum_{k=0}^\infty {\alpha\choose k}x^k$.
  • Knowledge of infinite series and their convergence criteria.
  • Basic grasp of probability theory, particularly the significance of the parameter $p$ in probabilistic models.
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  • Study the derivation of the binomial series and its applications in combinatorics.
  • Explore the properties of binomial coefficients and their role in series expansions.
  • Investigate convergence tests for infinite series to understand when such summations are valid.
  • Learn about the implications of the parameter $p$ in statistical distributions and its effect on series behavior.
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I came across some summation but have no idea how to simplify it.

$\sum_{x=0}^{\infty} \binom{x+r-2}{r-2}(1-p)^{x}=p^{1-r}$

Why is it so?
 
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pp123123 said:
I came across some summation but have no idea how to simplify it.

$\sum_{x=0}^{\infty} \binom{x+r-2}{r-2}(1-p)^{x}=p^{1-r}$

Why is it so?
Hint: Use the binomial series $$(1+x)^\alpha = \sum_{k=0}^\infty {\alpha\choose k}x^k,$$ with $x = -(1-p)$ and $\alpha = 1-r.$
 

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