Let \(a,b\in \mathbb{R}\). Then, for all \(n\in \mathbb{N}\),
\begin{equation*} (a+b)^n=\sum_{i=0}^n \binom{n}{i} a^{i}b^{n-i}. \end{equation*}
Remark
- For \(a,b\ge 0\) and \(k\in \mathbb{N}\) \[ (a+b)^k\le 2^k(a^k+b^k) \] (see (0x68bbccca) ).