Summing up the Series: $\sum^{n}_{x=2}x(x-1)\binom{n}{x}p^{x}q^{n-x}$

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Discussion Overview

The discussion revolves around the series $\sum^{n}_{x=2}x(x-1)\binom{n}{x}p^{x}q^{n-x}$, focusing on methods to find its sum. Participants explore connections to expected values and combinatorial simplifications.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant inquires about finding the sum of the series and suggests a connection to expected values.
  • Another participant hints at simplifying the term $x(x-1) \binom{n}{x}$ to facilitate the summation.
  • A participant reports a simplification leading to the expression $n(n-1)\sum^{n}_{x=2}{{n-2}\choose{x-2}}p^{x}q^{n-x}$ and mentions factoring out $p^{2}$ from the sum.
  • There is an assumption made regarding the condition $p + q = 1$ in the context of the discussion.

Areas of Agreement / Disagreement

Participants appear to agree on the approach to simplifying the series, but the overall resolution of the sum remains unclear, as no consensus on the final answer is explicitly stated.

Contextual Notes

The discussion does not clarify the assumptions underlying the simplifications or the conditions for the variables involved.

Who May Find This Useful

Individuals interested in combinatorial mathematics, series summation techniques, and expected value calculations may find this discussion relevant.

hanboning
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The series in question is \sum^{n}_{x=2}x(x-1)\binom{n}{x}p^{x}q^{n-x}. How do I find the sum of this?

I think this is related to expected values and doing cancelling with the x(x-1) with the combination.

Thanks!
 
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welcome to pf!

hi hanboning! welcome to pf! :smile:

hint: simplify x(x-1) nCx :wink:
 
Hi tim, thanks for your help. I might have gotten it.

So I simplified the things you said in your hint and I got n(n-1)\sum^{n}_{x=2}{{n-2}\choose{x-2}}p^{x}q^{n-x}. Then I moved a p^{2} out of the sum and got n(n-1)p^{2} as the answer.
 
s'right! :biggrin:

(i'm assuming p + q = 1? :wink:)
 
Yep, I appreciate your help. Happy new year.
 

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