Mathematical Analysis Fundamentals

Proof

Equality (a) is exactly the binomial formula:

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

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

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

In a similar way, using $k^2=k+k(k-1)$, we obtain

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

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