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Proof

Equality (a) is exactly the binomial formula:

$\sum _{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}{x}^{k}{\left(1-x\right)}^{n-k}={\left(x+1-x\right)}^{n}=1\text{.}$

Equality (b) can be reduced to the binomial formula as well:

$\begin{array}{cc}\hfill \sum _{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}{x}^{k}{\left(1-x\right)}^{n-k}k=& \mathit{nx}\sum _{k=1}^{n}\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k\right)!}{x}^{k-1}{\left(1-x\right)}^{n-k}\hfill \\ \hfill =& \mathit{nx}\sum _{k=0}^{n-1}\frac{\left(n-1\right)!}{k!\left(n-k-1\right)!}{x}^{k}{\left(1-x\right)}^{n-k-1}\hfill \\ \hfill =& \mathit{nx}{\left(x+1-x\right)}^{n-1}=\mathit{nx}\text{.}\hfill \end{array}$

In a similar way, using ${k}^{2}=k+k\left(k-1\right)$, we obtain

$\begin{array}{cc}\hfill \sum _{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}{x}^{k}{\left(1-x\right)}^{n-k}{k}^{2}& =\sum _{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}{x}^{k}{\left(1-x\right)}^{n-k}k\hfill \\ +\sum _{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}{x}^{k}{\left(1-x\right)}^{n-k}k\left(k-1\right)\text{.}\hfill \end{array}$

Here the first term is equal to $\mathit{nx}$ by ...

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