(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) :

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

Thank you to the contributors under CC BY-SA 4.0!