展开 $$$\left(a + b\right)^{6}$$$

展开 $$$\left(a + b\right)^{6}$$$。

该展开由以下公式给出：$$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$，其中$$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$和$$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$。

我们有$$$a = a$$$、$$$b = b$$$和$$$n = 6$$$。

因此，$$$\left(a + b\right)^{6} = \sum_{k=0}^{6} {\binom{6}{k}} a^{6 - k} b^{k}$$$。

现在，计算 $$$k$$$ 从 $$$0$$$ 到 $$$6$$$ 的每个取值对应的乘积。

$$$k = 0$$$: $$${\binom{6}{0}} a^{6 - 0} b^{0} = \frac{6!}{\left(6 - 0\right)! 0!} a^{6 - 0} b^{0} = a^{6}$$$

$$$k = 1$$$: $$${\binom{6}{1}} a^{6 - 1} b^{1} = \frac{6!}{\left(6 - 1\right)! 1!} a^{6 - 1} b^{1} = 6 a^{5} b$$$

$$$k = 2$$$: $$${\binom{6}{2}} a^{6 - 2} b^{2} = \frac{6!}{\left(6 - 2\right)! 2!} a^{6 - 2} b^{2} = 15 a^{4} b^{2}$$$

$$$k = 3$$$: $$${\binom{6}{3}} a^{6 - 3} b^{3} = \frac{6!}{\left(6 - 3\right)! 3!} a^{6 - 3} b^{3} = 20 a^{3} b^{3}$$$

$$$k = 4$$$: $$${\binom{6}{4}} a^{6 - 4} b^{4} = \frac{6!}{\left(6 - 4\right)! 4!} a^{6 - 4} b^{4} = 15 a^{2} b^{4}$$$

$$$k = 5$$$: $$${\binom{6}{5}} a^{6 - 5} b^{5} = \frac{6!}{\left(6 - 5\right)! 5!} a^{6 - 5} b^{5} = 6 a b^{5}$$$

$$$k = 6$$$: $$${\binom{6}{6}} a^{6 - 6} b^{6} = \frac{6!}{\left(6 - 6\right)! 6!} a^{6 - 6} b^{6} = b^{6}$$$

因此，$$$\left(a + b\right)^{6} = a^{6} + 6 a^{5} b + 15 a^{4} b^{2} + 20 a^{3} b^{3} + 15 a^{2} b^{4} + 6 a b^{5} + b^{6}$$$。

$$$\left(a + b\right)^{6} = a^{6} + 6 a^{5} b + 15 a^{4} b^{2} + 20 a^{3} b^{3} + 15 a^{2} b^{4} + 6 a b^{5} + b^{6}$$$A