Is the Summation of the Binomial Theorem Equal to Zero?

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Discussion Overview

The discussion revolves around the summation of the binomial theorem and whether the expression \(\sum_{k=0}^{n}{n\choose k}\,(-1)^k\) equals zero for all \(n\), particularly examining the case when \(n = 0\).

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant states that substituting \(x=1\) and \(y=-1\) into the binomial theorem leads to the expression summing to zero, suggesting that \((1-1)^n=0\) implies the summation equals zero.
  • Another participant agrees with the assertion that the summation equals zero, but notes that this is not the case when \(n=0\).
  • A follow-up question is posed regarding the value of the summation when \(n=0\), implying it may yield a different result.
  • A further contribution points out that \(0^0\) is not well-defined and raises concerns about the definition of binomial coefficients for \(k \neq 0\) when \(n=0\).

Areas of Agreement / Disagreement

Participants generally agree that the summation equals zero for \(n > 0\), but there is uncertainty and disagreement regarding the case when \(n = 0.

Contextual Notes

The discussion highlights limitations in the definitions and interpretations of \(0^0\) and binomial coefficients, particularly in edge cases.

EngWiPy
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Hello,

All we know the Binomial theorem which may be stated mathematically as:

\left(x+y\right)^n=\sum_{k=0}^n{n\choose k}y^k\,x^{n-k}

Now suppose that we have the following mathematical expression:

\sum_{k=0}^{n}{n\choose k}\,(-1)^k

if we substitute x=1 and y=-1 in the first equation we get the second. Is that mean the second equation is essentially zero, since (1-1)^n=0??

Regards
 
Mathematics news on Phys.org
Yes, indeed, unless n = 0.
 
Moo Of Doom said:
Yes, indeed, unless n = 0.

Why? In the case that n = 0, what will be the answer? 1?
 
00 is not well-defined and neither is 0Ck for any k <> 0 (although there are generalizations that extend the domain beyond the definition using just factorials).
 

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