二项式定理

对于 $a,\, b \in \mathbb{R}$ 和 $n \in \mathbb{N}^*$，有

$$\left(a + b\right)^n = \sum_{i = 0}^n \binom{n}{i} a^{n - i} b^i$$

证明

1. 当 $n = 1$ 时，显然结论成立
2. 当 $n = k$ 时，假设结论成立，有

$$\left(a + b\right)^k = \sum_{i = 0}^k \binom{k}{i} a^{k - i} b^i$$

当 $n = k + 1$ 时，有

$$\begin{aligned} \left(a + b\right)^{k + 1} &= \left(a + b\right)^k \left(a + b\right) \\ &= \left(\sum_{i = 0}^k \binom{k}{i} a^{k - i} b^i\right) \left(a + b\right) \\ &= \sum_{i = 0}^k \binom{k}{i} a^{k - i + 1} b^i + \sum_{i = 0}^k \binom{k}{i} a^{k - i} b^{i + 1} \\ &= \sum_{i = 1}^{k + 1} \binom{k}{i - 1} a^{k - i + 1} b^{i - 1} + \sum_{i = 0}^k \binom{k}{i} a^{k - i} b^{i + 1} \\ &= a^{k+1} + \sum_{i = 1}^k \binom{k}{i - 1} a^{k - i + 1} b^i + \sum_{i = 1}^k \binom{k}{i} a^{k - i + 1} b^i + b^{k + 1} \\ &= a^{k+1} + \sum_{i = 1}^k \left(\binom{k}{i - 1} + \binom{k}{i}\right) a^{k - i + 1} b^i + b^{k + 1} \\ &= a^{k+1} + \sum_{i = 1}^k \binom{k + 1}{i} a^{k - i + 1} b^i + b^{k + 1} \\ &= \sum_{i = 0}^{k + 1} \binom{k + 1}{i} a^{k + 1 - i} b^i \end{aligned}$$

结论仍成立，综合 (1) 和 (2)，结论成立。