# Proof

(related to Proposition: Sum of Binomial Coefficients II)

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 following mathematical concepts and theorems: * a binomial identity for $$k$$ times $$n$$ choose $$k$$ , * some rules of calculation in a ring, and * the sum of binomial coefficients (i) :

$\begin{array}{rcll} \sum_{k=0}^nk\binom nk(1-x)^{n-k}x^k&=&\sum_{k=1}^nk\binom nk(1-x)^{n-k}x^k&\text{first summand can be ignored (multiplication by 0)}\\ &=&\sum_{k=1}^nn\binom{n-1}{k-1}(1-x)^{n-k}x^k&\text{binomial identity for k times n choose k}\\ &=&n\sum_{k=1}^n\binom{n-1}{k-1}(1-x)^{n-k}x^k&\text{distributivity law in the ring }R\\ &=&n\sum_{k=0}^{n-1}\binom{n-1}{k}(1-x)^{(n-1)-k}x^{k+1}&\text{change of summation index }k\to k-1\\ &=&nx\sum_{k=0}^{n-1}\binom{n-1}{k}(1-x)^{(n-1)-k}x^k&\text{distributivity law in the ring }R\\ &=&nx\cdot 1&\text{sum of binomial coefficients (i)}\\ &=&nx&\text{existence of identity in }R\\ \end{array}$

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