Proof

(related to Proposition: Sum of Binomial Coefficients I)

Let \(x\) be an element of a ring \(x\in(R,+,\cdot) \) and let \(n\ge 1\) be a natural number. In the following calculation, we use the binomial theorem and some rules of calculation in a ring:

\[ \begin{array}{rcll} \sum_{k=0}^n\binom nk(1-x)^{n-k}x^k&=&((1-x)+x)^{n}&\text{binomial theorem}\\ &=&(1-x+x)^{n}&\text{associativity of addition in }R\\ &=&(1+0)^{n}&\text{existence of negative inverse in }R\\ &=&1^{n}&\text{existence of zero in }R\\ &=&1&\text{existence of identity in }R\\ \end{array} \]


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References

Bibliography

  1. Bosch, Karl: "Elementare Einf├╝hrung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition