Multipliziere $$$\left(x + y\right)^{6}$$$ aus

Multipliziere $$$\left(x + y\right)^{6}$$$ aus.

Die Entwicklung wird durch die folgende Formel gegeben: $$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$, wobei $$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$ und $$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$.

Es gilt, dass $$$a = x$$$, $$$b = y$$$ und $$$n = 6$$$.

Daher $$$\left(x + y\right)^{6} = \sum_{k=0}^{6} {\binom{6}{k}} x^{6 - k} y^{k}$$$.

Berechne nun das Produkt für jeden Wert von $$$k$$$ von $$$0$$$ bis $$$6$$$.

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

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

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

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

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

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

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

Somit gilt $$$\left(x + y\right)^{6} = x^{6} + 6 x^{5} y + 15 x^{4} y^{2} + 20 x^{3} y^{3} + 15 x^{2} y^{4} + 6 x y^{5} + y^{6}$$$.

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