Mathematical Analysis Fundamentals

Proof

We will use the induction principle. Fix arbitrary $a\text{,}b\in \mathbb{R}$ and let the preceding equality be the assertion $P(n)$. Then $P(1)$ is obviously true. Assume that $P(n)$ is true for some $n$. Then by Lemma 2.11,

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